A kind of asymptote for generalized alternating series












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This is an interesting problem (Problem 1.6.1) from the book "Problems in Real Analysis" by Radulescu et al.:




Let $sum b_n a_n$ be a convergent series, where $a_1 geq a_2 geq ldots geq 0$, and $b_i = pm 1$ for all $i$.



Show that $(b_1 + b_2 + ldots + b_n)a_n$ converges to $0$ as $n to infty$.




I'd love to see some solutions.






real-analysis







share|cite|improve this question



























    0














    This is an interesting problem (Problem 1.6.1) from the book "Problems in Real Analysis" by Radulescu et al.:




    Let $sum b_n a_n$ be a convergent series, where $a_1 geq a_2 geq ldots geq 0$, and $b_i = pm 1$ for all $i$.



    Show that $(b_1 + b_2 + ldots + b_n)a_n$ converges to $0$ as $n to infty$.




    I'd love to see some solutions.






    real-analysis







    share|cite|improve this question

























      0












      0








      0







      This is an interesting problem (Problem 1.6.1) from the book "Problems in Real Analysis" by Radulescu et al.:




      Let $sum b_n a_n$ be a convergent series, where $a_1 geq a_2 geq ldots geq 0$, and $b_i = pm 1$ for all $i$.



      Show that $(b_1 + b_2 + ldots + b_n)a_n$ converges to $0$ as $n to infty$.




      I'd love to see some solutions.






      real-analysis







      share|cite|improve this question













      This is an interesting problem (Problem 1.6.1) from the book "Problems in Real Analysis" by Radulescu et al.:




      Let $sum b_n a_n$ be a convergent series, where $a_1 geq a_2 geq ldots geq 0$, and $b_i = pm 1$ for all $i$.



      Show that $(b_1 + b_2 + ldots + b_n)a_n$ converges to $0$ as $n to infty$.




      I'd love to see some solutions.






      real-analysis




      real-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 29 '18 at 12:09









      user97678

      756




      756






















          1 Answer
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          There is a simple proof, without to many technical steps, with the help of Kronecker's lemma. In fact, it is a special case of Kronecker's lemma.




          If $alpha_k in mathbb{R}$ and $(beta_k)_k$ is an increasing sequence of positive real numbers with $beta_k rightarrow infty$ such that $sum_{k=1}^infty alpha_k$ is convergent, then $frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k rightarrow 0$ for $n rightarrow infty$.




          Proof: Write $S_n := sum_{k=1}^n alpha_k$, then we have
          $$frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k = frac{1}{beta_n} sum_{k=1}^n beta_k (S_k -S_{k-1}) = S_n - frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k $$
          Since $S_n rightarrow s$, it remains to prove that $ frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k rightarrow s$. Take $|S_n -s| < varepsilon$ for all $n ge N$. Then we have for all $n > N$ that
          begin{align}
          &frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k-(1- beta_1/beta_n)s \
          =& frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) (S_k-s) \
          =&frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) + frac{1}{beta_n} sum_{k=N}^{n-1} (beta_{k+1}-beta_k) (S_k-s)
          end{align}
          We have $(1- beta_1/beta_n)s rightarrow s$ and $frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) rightarrow 0$ for fixed $N$. On the other hand
          $$left|frac{1}{beta_n} sum_{k=N}^{n} (beta_{k+1}-beta_k) (S_k-s) right| le varepsilon frac{1}{beta_n} (beta_n - beta_N) le varepsilon M $$
          This proves Kronecker's lemma.



          Now, we would like to apply the lemma as follows. Our first observation is that $a_n rightarrow 0$ for $n rightarrow infty$, because $|a_n b_b| = |a_n|$ and $b_n a_n rightarrow 0$ for $n rightarrow infty$, since the sum converges. Thus, we can take $beta_n = a_k^{-1}$ and this sequence satisfies Kronecker's lemma, because $(a_n)_n$ is also decreasing. Since $sum_{k=1}^infty (b_k a_k)$ is convergent (we take $alpha_k = b_k a_k$), we get that
          $$a_m sum_{k=1}^m b_k = frac{1}{beta_m} sum_{k=1}^m alpha_k beta_k rightarrow 0.$$






          share|cite|improve this answer























          • Why does a $k$ fulfilling your inequality (2) exist?
            – user97678
            Nov 29 '18 at 13:40










          • We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
            – p4sch
            Nov 29 '18 at 13:44












          • I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
            – user97678
            Nov 29 '18 at 13:48










          • Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
            – p4sch
            Nov 29 '18 at 13:59












          • But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
            – user97678
            Nov 29 '18 at 14:01











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          2














          There is a simple proof, without to many technical steps, with the help of Kronecker's lemma. In fact, it is a special case of Kronecker's lemma.




          If $alpha_k in mathbb{R}$ and $(beta_k)_k$ is an increasing sequence of positive real numbers with $beta_k rightarrow infty$ such that $sum_{k=1}^infty alpha_k$ is convergent, then $frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k rightarrow 0$ for $n rightarrow infty$.




          Proof: Write $S_n := sum_{k=1}^n alpha_k$, then we have
          $$frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k = frac{1}{beta_n} sum_{k=1}^n beta_k (S_k -S_{k-1}) = S_n - frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k $$
          Since $S_n rightarrow s$, it remains to prove that $ frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k rightarrow s$. Take $|S_n -s| < varepsilon$ for all $n ge N$. Then we have for all $n > N$ that
          begin{align}
          &frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k-(1- beta_1/beta_n)s \
          =& frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) (S_k-s) \
          =&frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) + frac{1}{beta_n} sum_{k=N}^{n-1} (beta_{k+1}-beta_k) (S_k-s)
          end{align}
          We have $(1- beta_1/beta_n)s rightarrow s$ and $frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) rightarrow 0$ for fixed $N$. On the other hand
          $$left|frac{1}{beta_n} sum_{k=N}^{n} (beta_{k+1}-beta_k) (S_k-s) right| le varepsilon frac{1}{beta_n} (beta_n - beta_N) le varepsilon M $$
          This proves Kronecker's lemma.



          Now, we would like to apply the lemma as follows. Our first observation is that $a_n rightarrow 0$ for $n rightarrow infty$, because $|a_n b_b| = |a_n|$ and $b_n a_n rightarrow 0$ for $n rightarrow infty$, since the sum converges. Thus, we can take $beta_n = a_k^{-1}$ and this sequence satisfies Kronecker's lemma, because $(a_n)_n$ is also decreasing. Since $sum_{k=1}^infty (b_k a_k)$ is convergent (we take $alpha_k = b_k a_k$), we get that
          $$a_m sum_{k=1}^m b_k = frac{1}{beta_m} sum_{k=1}^m alpha_k beta_k rightarrow 0.$$






          share|cite|improve this answer























          • Why does a $k$ fulfilling your inequality (2) exist?
            – user97678
            Nov 29 '18 at 13:40










          • We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
            – p4sch
            Nov 29 '18 at 13:44












          • I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
            – user97678
            Nov 29 '18 at 13:48










          • Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
            – p4sch
            Nov 29 '18 at 13:59












          • But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
            – user97678
            Nov 29 '18 at 14:01
















          2














          There is a simple proof, without to many technical steps, with the help of Kronecker's lemma. In fact, it is a special case of Kronecker's lemma.




          If $alpha_k in mathbb{R}$ and $(beta_k)_k$ is an increasing sequence of positive real numbers with $beta_k rightarrow infty$ such that $sum_{k=1}^infty alpha_k$ is convergent, then $frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k rightarrow 0$ for $n rightarrow infty$.




          Proof: Write $S_n := sum_{k=1}^n alpha_k$, then we have
          $$frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k = frac{1}{beta_n} sum_{k=1}^n beta_k (S_k -S_{k-1}) = S_n - frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k $$
          Since $S_n rightarrow s$, it remains to prove that $ frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k rightarrow s$. Take $|S_n -s| < varepsilon$ for all $n ge N$. Then we have for all $n > N$ that
          begin{align}
          &frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k-(1- beta_1/beta_n)s \
          =& frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) (S_k-s) \
          =&frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) + frac{1}{beta_n} sum_{k=N}^{n-1} (beta_{k+1}-beta_k) (S_k-s)
          end{align}
          We have $(1- beta_1/beta_n)s rightarrow s$ and $frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) rightarrow 0$ for fixed $N$. On the other hand
          $$left|frac{1}{beta_n} sum_{k=N}^{n} (beta_{k+1}-beta_k) (S_k-s) right| le varepsilon frac{1}{beta_n} (beta_n - beta_N) le varepsilon M $$
          This proves Kronecker's lemma.



          Now, we would like to apply the lemma as follows. Our first observation is that $a_n rightarrow 0$ for $n rightarrow infty$, because $|a_n b_b| = |a_n|$ and $b_n a_n rightarrow 0$ for $n rightarrow infty$, since the sum converges. Thus, we can take $beta_n = a_k^{-1}$ and this sequence satisfies Kronecker's lemma, because $(a_n)_n$ is also decreasing. Since $sum_{k=1}^infty (b_k a_k)$ is convergent (we take $alpha_k = b_k a_k$), we get that
          $$a_m sum_{k=1}^m b_k = frac{1}{beta_m} sum_{k=1}^m alpha_k beta_k rightarrow 0.$$






          share|cite|improve this answer























          • Why does a $k$ fulfilling your inequality (2) exist?
            – user97678
            Nov 29 '18 at 13:40










          • We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
            – p4sch
            Nov 29 '18 at 13:44












          • I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
            – user97678
            Nov 29 '18 at 13:48










          • Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
            – p4sch
            Nov 29 '18 at 13:59












          • But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
            – user97678
            Nov 29 '18 at 14:01














          2












          2








          2






          There is a simple proof, without to many technical steps, with the help of Kronecker's lemma. In fact, it is a special case of Kronecker's lemma.




          If $alpha_k in mathbb{R}$ and $(beta_k)_k$ is an increasing sequence of positive real numbers with $beta_k rightarrow infty$ such that $sum_{k=1}^infty alpha_k$ is convergent, then $frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k rightarrow 0$ for $n rightarrow infty$.




          Proof: Write $S_n := sum_{k=1}^n alpha_k$, then we have
          $$frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k = frac{1}{beta_n} sum_{k=1}^n beta_k (S_k -S_{k-1}) = S_n - frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k $$
          Since $S_n rightarrow s$, it remains to prove that $ frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k rightarrow s$. Take $|S_n -s| < varepsilon$ for all $n ge N$. Then we have for all $n > N$ that
          begin{align}
          &frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k-(1- beta_1/beta_n)s \
          =& frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) (S_k-s) \
          =&frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) + frac{1}{beta_n} sum_{k=N}^{n-1} (beta_{k+1}-beta_k) (S_k-s)
          end{align}
          We have $(1- beta_1/beta_n)s rightarrow s$ and $frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) rightarrow 0$ for fixed $N$. On the other hand
          $$left|frac{1}{beta_n} sum_{k=N}^{n} (beta_{k+1}-beta_k) (S_k-s) right| le varepsilon frac{1}{beta_n} (beta_n - beta_N) le varepsilon M $$
          This proves Kronecker's lemma.



          Now, we would like to apply the lemma as follows. Our first observation is that $a_n rightarrow 0$ for $n rightarrow infty$, because $|a_n b_b| = |a_n|$ and $b_n a_n rightarrow 0$ for $n rightarrow infty$, since the sum converges. Thus, we can take $beta_n = a_k^{-1}$ and this sequence satisfies Kronecker's lemma, because $(a_n)_n$ is also decreasing. Since $sum_{k=1}^infty (b_k a_k)$ is convergent (we take $alpha_k = b_k a_k$), we get that
          $$a_m sum_{k=1}^m b_k = frac{1}{beta_m} sum_{k=1}^m alpha_k beta_k rightarrow 0.$$






          share|cite|improve this answer














          There is a simple proof, without to many technical steps, with the help of Kronecker's lemma. In fact, it is a special case of Kronecker's lemma.




          If $alpha_k in mathbb{R}$ and $(beta_k)_k$ is an increasing sequence of positive real numbers with $beta_k rightarrow infty$ such that $sum_{k=1}^infty alpha_k$ is convergent, then $frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k rightarrow 0$ for $n rightarrow infty$.




          Proof: Write $S_n := sum_{k=1}^n alpha_k$, then we have
          $$frac{1}{beta_n} sum_{k=1}^n beta_k alpha_k = frac{1}{beta_n} sum_{k=1}^n beta_k (S_k -S_{k-1}) = S_n - frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k $$
          Since $S_n rightarrow s$, it remains to prove that $ frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k rightarrow s$. Take $|S_n -s| < varepsilon$ for all $n ge N$. Then we have for all $n > N$ that
          begin{align}
          &frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) S_k-(1- beta_1/beta_n)s \
          =& frac{1}{beta_n} sum_{k=1}^{n-1} (beta_{k+1}-beta_k) (S_k-s) \
          =&frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) + frac{1}{beta_n} sum_{k=N}^{n-1} (beta_{k+1}-beta_k) (S_k-s)
          end{align}
          We have $(1- beta_1/beta_n)s rightarrow s$ and $frac{1}{beta_n} sum_{k=1}^{N-1} (beta_{k+1}-beta_k) (S_k-s) rightarrow 0$ for fixed $N$. On the other hand
          $$left|frac{1}{beta_n} sum_{k=N}^{n} (beta_{k+1}-beta_k) (S_k-s) right| le varepsilon frac{1}{beta_n} (beta_n - beta_N) le varepsilon M $$
          This proves Kronecker's lemma.



          Now, we would like to apply the lemma as follows. Our first observation is that $a_n rightarrow 0$ for $n rightarrow infty$, because $|a_n b_b| = |a_n|$ and $b_n a_n rightarrow 0$ for $n rightarrow infty$, since the sum converges. Thus, we can take $beta_n = a_k^{-1}$ and this sequence satisfies Kronecker's lemma, because $(a_n)_n$ is also decreasing. Since $sum_{k=1}^infty (b_k a_k)$ is convergent (we take $alpha_k = b_k a_k$), we get that
          $$a_m sum_{k=1}^m b_k = frac{1}{beta_m} sum_{k=1}^m alpha_k beta_k rightarrow 0.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 29 '18 at 15:59

























          answered Nov 29 '18 at 12:57









          p4sch

          4,770217




          4,770217












          • Why does a $k$ fulfilling your inequality (2) exist?
            – user97678
            Nov 29 '18 at 13:40










          • We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
            – p4sch
            Nov 29 '18 at 13:44












          • I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
            – user97678
            Nov 29 '18 at 13:48










          • Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
            – p4sch
            Nov 29 '18 at 13:59












          • But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
            – user97678
            Nov 29 '18 at 14:01


















          • Why does a $k$ fulfilling your inequality (2) exist?
            – user97678
            Nov 29 '18 at 13:40










          • We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
            – p4sch
            Nov 29 '18 at 13:44












          • I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
            – user97678
            Nov 29 '18 at 13:48










          • Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
            – p4sch
            Nov 29 '18 at 13:59












          • But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
            – user97678
            Nov 29 '18 at 14:01
















          Why does a $k$ fulfilling your inequality (2) exist?
          – user97678
          Nov 29 '18 at 13:40




          Why does a $k$ fulfilling your inequality (2) exist?
          – user97678
          Nov 29 '18 at 13:40












          We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
          – p4sch
          Nov 29 '18 at 13:44






          We have $n_k rightarrow infty$ for $k rightarrow infty$. On the other hand, we can use that the series convegences, i.e. is a Cauchy-sequence. Thus, for any $varepsilon >0$ there exists $N in mathbb{N}$ with $|sum_{i=n}^m b_i a_i| < varepsilon$ for all $m >n ge N$. Specially, we can take (for $k$ large enough) $n = n_k$ and $m = (n_{k+1}-1) wedge m$.
          – p4sch
          Nov 29 '18 at 13:44














          I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
          – user97678
          Nov 29 '18 at 13:48




          I can see why due to Cauchy it has to be $< epsilon$, but why is the left part of the inequality correct? I'm not sure why $n_{k+1} wedge (m-n_k)a_m$ has to be smaller than the sum
          – user97678
          Nov 29 '18 at 13:48












          Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
          – p4sch
          Nov 29 '18 at 13:59






          Note that $(a_n)$ is decreasing and that we have $b_i =1$ or $b_i = -1$ for all terms in the sum. Thus the sum is just $left| sum_{i=n_l}^{(n_{l+1}-1) wedge m} a_i right| ge a_m [(n_{l+1}-1) wedge m-n_l+1]$.
          – p4sch
          Nov 29 '18 at 13:59














          But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
          – user97678
          Nov 29 '18 at 14:01




          But why is $b_i$'s sign fixed for all terms in the sum? Isn't your $m$ unbounded? You can pick it high enough that the $b_i$s start changing signs, no?
          – user97678
          Nov 29 '18 at 14:01


















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