Is there any formula to find the sum of a series like $ (n/16)^{3/4}+(n/16^2)^{3/4}+(n/16^3)^{3/4} cdots $
I am wondering if there is any way to find the sum of series like this either finite(upto some n and we do not know the value of n) or infinite.
$$left(frac{n}{16}right)^{3/4}+left(frac{n}{16^2}right)^{3/4}+left(frac{n}{16^3}right)^{3/4}+ldots$$
sequences-and-series
edited Oct 24 '18 at 21:57
rtybase
10.5k21533
asked Oct 16 '18 at 6:40
SreeSree
111
add a comment |
I am wondering if there is any way to find the sum of series like this either finite(upto some n and we do not know the value of n) or infinite.
$$left(frac{n}{16}right)^{3/4}+left(frac{n}{16^2}right)^{3/4}+left(frac{n}{16^3}right)^{3/4}+ldots$$
sequences-and-series
edited Oct 24 '18 at 21:57
rtybase
10.5k21533
asked Oct 16 '18 at 6:40
SreeSree
111
2
You want to calculate $$S = n^{frac{3}{4}}sum_{i=1}^inftyfrac{1}{16^{frac{3i}{4}}}text{?}$$ Please take a look at the geometric series...
– denklo
Oct 16 '18 at 6:45
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Oct 16 '18 at 6:50
add a comment |
I am wondering if there is any way to find the sum of series like this either finite(upto some n and we do not know the value of n) or infinite.
$$left(frac{n}{16}right)^{3/4}+left(frac{n}{16^2}right)^{3/4}+left(frac{n}{16^3}right)^{3/4}+ldots$$
sequences-and-series
edited Oct 24 '18 at 21:57
rtybase
10.5k21533
asked Oct 16 '18 at 6:40
SreeSree
111
I am wondering if there is any way to find the sum of series like this either finite(upto some n and we do not know the value of n) or infinite.
$$left(frac{n}{16}right)^{3/4}+left(frac{n}{16^2}right)^{3/4}+left(frac{n}{16^3}right)^{3/4}+ldots$$
sequences-and-series
sequences-and-series
edited Oct 24 '18 at 21:57
rtybase
10.5k21533
asked Oct 16 '18 at 6:40
SreeSree
111
edited Oct 24 '18 at 21:57
rtybase
10.5k21533
asked Oct 16 '18 at 6:40
SreeSree
111
edited Oct 24 '18 at 21:57
rtybase
10.5k21533
edited Oct 24 '18 at 21:57
rtybase
10.5k21533
edited Oct 24 '18 at 21:57
rtybase
10.5k21533
10.5k21533
asked Oct 16 '18 at 6:40
SreeSree
111
asked Oct 16 '18 at 6:40
SreeSree
111
asked Oct 16 '18 at 6:40
SreeSree
111
111
2
You want to calculate $$S = n^{frac{3}{4}}sum_{i=1}^inftyfrac{1}{16^{frac{3i}{4}}}text{?}$$ Please take a look at the geometric series...
– denklo
Oct 16 '18 at 6:45
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Oct 16 '18 at 6:50
add a comment |
2
You want to calculate $$S = n^{frac{3}{4}}sum_{i=1}^inftyfrac{1}{16^{frac{3i}{4}}}text{?}$$ Please take a look at the geometric series...
– denklo
Oct 16 '18 at 6:45
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Oct 16 '18 at 6:50
2
2
You want to calculate $$S = n^{frac{3}{4}}sum_{i=1}^inftyfrac{1}{16^{frac{3i}{4}}}text{?}$$ Please take a look at the geometric series...
– denklo
Oct 16 '18 at 6:45
You want to calculate $$S = n^{frac{3}{4}}sum_{i=1}^inftyfrac{1}{16^{frac{3i}{4}}}text{?}$$ Please take a look at the geometric series...
– denklo
Oct 16 '18 at 6:45
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Oct 16 '18 at 6:50
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Oct 16 '18 at 6:50
add a comment |
2 Answers
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$$left(frac{n}{16}right)^{3/4}+left(frac{n}{16^2}right)^{3/4}+left(frac{n}{16^3}right)^{3/4}+ldots{=n^{3over 4}sum_{k=1}^{infty}{1over 16^{3kover 4}}\=n^{3over 4}sum_{k=1}^{infty}{1over 2^{3k}}\=n^{3over 4}sum_{k=1}^{infty}{1over 8^{k}}=\=n^{3over 4}{{1over 8}over 1-{1over 8}}\={1over 7}n^{3over 4}}$$
answered Nov 30 '18 at 19:47
Mostafa AyazMostafa Ayaz
14.5k3937
add a comment |
Rewrite it in summation notation to see that$$S=n^{3/4}sumlimits_{kgeq1}16^{-3k/4}$$Now use the formula for the infinite geometric sequence for $|r|<1$
$$sumlimits_{ngeq1}r^n=frac r{1-r}$$This can be proven by expanding out the infinite sum and noting how$$begin{align*}A & =r+r^2+r^3+cdots\ Ar & =phantom{r+}spacespace r^2+r^3+cdotsend{align*}$$Now subtract and isolate $A$. In this case, $r=16^{-3/4}$ and simplifying, the answer turns out to be$$Scolor{blue}{=frac {n^{3/4}}{7}}$$
answered Nov 30 '18 at 19:56
Frank W.Frank W.
3,1891321
add a comment |
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2 Answers
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2 Answers
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$$left(frac{n}{16}right)^{3/4}+left(frac{n}{16^2}right)^{3/4}+left(frac{n}{16^3}right)^{3/4}+ldots{=n^{3over 4}sum_{k=1}^{infty}{1over 16^{3kover 4}}\=n^{3over 4}sum_{k=1}^{infty}{1over 2^{3k}}\=n^{3over 4}sum_{k=1}^{infty}{1over 8^{k}}=\=n^{3over 4}{{1over 8}over 1-{1over 8}}\={1over 7}n^{3over 4}}$$
answered Nov 30 '18 at 19:47
Mostafa AyazMostafa Ayaz
14.5k3937
add a comment |
$$left(frac{n}{16}right)^{3/4}+left(frac{n}{16^2}right)^{3/4}+left(frac{n}{16^3}right)^{3/4}+ldots{=n^{3over 4}sum_{k=1}^{infty}{1over 16^{3kover 4}}\=n^{3over 4}sum_{k=1}^{infty}{1over 2^{3k}}\=n^{3over 4}sum_{k=1}^{infty}{1over 8^{k}}=\=n^{3over 4}{{1over 8}over 1-{1over 8}}\={1over 7}n^{3over 4}}$$
answered Nov 30 '18 at 19:47
Mostafa AyazMostafa Ayaz
14.5k3937
add a comment |
$$left(frac{n}{16}right)^{3/4}+left(frac{n}{16^2}right)^{3/4}+left(frac{n}{16^3}right)^{3/4}+ldots{=n^{3over 4}sum_{k=1}^{infty}{1over 16^{3kover 4}}\=n^{3over 4}sum_{k=1}^{infty}{1over 2^{3k}}\=n^{3over 4}sum_{k=1}^{infty}{1over 8^{k}}=\=n^{3over 4}{{1over 8}over 1-{1over 8}}\={1over 7}n^{3over 4}}$$
answered Nov 30 '18 at 19:47
Mostafa AyazMostafa Ayaz
14.5k3937
$$left(frac{n}{16}right)^{3/4}+left(frac{n}{16^2}right)^{3/4}+left(frac{n}{16^3}right)^{3/4}+ldots{=n^{3over 4}sum_{k=1}^{infty}{1over 16^{3kover 4}}\=n^{3over 4}sum_{k=1}^{infty}{1over 2^{3k}}\=n^{3over 4}sum_{k=1}^{infty}{1over 8^{k}}=\=n^{3over 4}{{1over 8}over 1-{1over 8}}\={1over 7}n^{3over 4}}$$
answered Nov 30 '18 at 19:47
Mostafa AyazMostafa Ayaz
14.5k3937
answered Nov 30 '18 at 19:47
Mostafa AyazMostafa Ayaz
14.5k3937
answered Nov 30 '18 at 19:47
Mostafa AyazMostafa Ayaz
14.5k3937
answered Nov 30 '18 at 19:47
Mostafa AyazMostafa Ayaz
14.5k3937
14.5k3937
add a comment |
add a comment |
Rewrite it in summation notation to see that$$S=n^{3/4}sumlimits_{kgeq1}16^{-3k/4}$$Now use the formula for the infinite geometric sequence for $|r|<1$
$$sumlimits_{ngeq1}r^n=frac r{1-r}$$This can be proven by expanding out the infinite sum and noting how$$begin{align*}A & =r+r^2+r^3+cdots\ Ar & =phantom{r+}spacespace r^2+r^3+cdotsend{align*}$$Now subtract and isolate $A$. In this case, $r=16^{-3/4}$ and simplifying, the answer turns out to be$$Scolor{blue}{=frac {n^{3/4}}{7}}$$
answered Nov 30 '18 at 19:56
Frank W.Frank W.
3,1891321
add a comment |
Rewrite it in summation notation to see that$$S=n^{3/4}sumlimits_{kgeq1}16^{-3k/4}$$Now use the formula for the infinite geometric sequence for $|r|<1$
$$sumlimits_{ngeq1}r^n=frac r{1-r}$$This can be proven by expanding out the infinite sum and noting how$$begin{align*}A & =r+r^2+r^3+cdots\ Ar & =phantom{r+}spacespace r^2+r^3+cdotsend{align*}$$Now subtract and isolate $A$. In this case, $r=16^{-3/4}$ and simplifying, the answer turns out to be$$Scolor{blue}{=frac {n^{3/4}}{7}}$$
answered Nov 30 '18 at 19:56
Frank W.Frank W.
3,1891321
add a comment |
Rewrite it in summation notation to see that$$S=n^{3/4}sumlimits_{kgeq1}16^{-3k/4}$$Now use the formula for the infinite geometric sequence for $|r|<1$
$$sumlimits_{ngeq1}r^n=frac r{1-r}$$This can be proven by expanding out the infinite sum and noting how$$begin{align*}A & =r+r^2+r^3+cdots\ Ar & =phantom{r+}spacespace r^2+r^3+cdotsend{align*}$$Now subtract and isolate $A$. In this case, $r=16^{-3/4}$ and simplifying, the answer turns out to be$$Scolor{blue}{=frac {n^{3/4}}{7}}$$
answered Nov 30 '18 at 19:56
Frank W.Frank W.
3,1891321
Rewrite it in summation notation to see that$$S=n^{3/4}sumlimits_{kgeq1}16^{-3k/4}$$Now use the formula for the infinite geometric sequence for $|r|<1$
$$sumlimits_{ngeq1}r^n=frac r{1-r}$$This can be proven by expanding out the infinite sum and noting how$$begin{align*}A & =r+r^2+r^3+cdots\ Ar & =phantom{r+}spacespace r^2+r^3+cdotsend{align*}$$Now subtract and isolate $A$. In this case, $r=16^{-3/4}$ and simplifying, the answer turns out to be$$Scolor{blue}{=frac {n^{3/4}}{7}}$$
answered Nov 30 '18 at 19:56
Frank W.Frank W.
3,1891321
answered Nov 30 '18 at 19:56
Frank W.Frank W.
3,1891321
answered Nov 30 '18 at 19:56
Frank W.Frank W.
3,1891321
answered Nov 30 '18 at 19:56
Frank W.Frank W.
3,1891321
3,1891321
add a comment |
add a comment |
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You want to calculate $$S = n^{frac{3}{4}}sum_{i=1}^inftyfrac{1}{16^{frac{3i}{4}}}text{?}$$ Please take a look at the geometric series...
– denklo
Oct 16 '18 at 6:45
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Oct 16 '18 at 6:50