Method for determining convergence of sinus/cosinus series












-2














What's the method used for testing convergence of these type of series?
$$
sum_{n = 2}^inftysinfrac{(n^2-1)pi}{n}
$$










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    -2














    What's the method used for testing convergence of these type of series?
    $$
    sum_{n = 2}^inftysinfrac{(n^2-1)pi}{n}
    $$










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      -2












      -2








      -2







      What's the method used for testing convergence of these type of series?
      $$
      sum_{n = 2}^inftysinfrac{(n^2-1)pi}{n}
      $$










      share|cite|improve this question















      What's the method used for testing convergence of these type of series?
      $$
      sum_{n = 2}^inftysinfrac{(n^2-1)pi}{n}
      $$







      calculus sequences-and-series






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      edited Nov 30 '18 at 12:50









      Arthur

      111k7105186




      111k7105186










      asked Nov 30 '18 at 12:12









      sydsyd

      52




      52






















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          You have



          $$sinleft(frac{(n^2-1)pi}{n}right) = sinleft(npi -frac{pi}{n}right) $$



          $$=sin(npi)cos(pi/n) - cos(npi)sin(pi/n)$$



          $$= 0 +(-1)^{n+1}sin(pi/n).$$



          So on one hand you have an alternating series and on the other hand you have something asymptotic to the harmonic series.






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          • Note that the convergence follows by the Leibniz criterion. :-)
            – p4sch
            Nov 30 '18 at 13:41











          Your Answer





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          1 Answer
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          1 Answer
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          You have



          $$sinleft(frac{(n^2-1)pi}{n}right) = sinleft(npi -frac{pi}{n}right) $$



          $$=sin(npi)cos(pi/n) - cos(npi)sin(pi/n)$$



          $$= 0 +(-1)^{n+1}sin(pi/n).$$



          So on one hand you have an alternating series and on the other hand you have something asymptotic to the harmonic series.






          share|cite|improve this answer





















          • Note that the convergence follows by the Leibniz criterion. :-)
            – p4sch
            Nov 30 '18 at 13:41
















          1














          You have



          $$sinleft(frac{(n^2-1)pi}{n}right) = sinleft(npi -frac{pi}{n}right) $$



          $$=sin(npi)cos(pi/n) - cos(npi)sin(pi/n)$$



          $$= 0 +(-1)^{n+1}sin(pi/n).$$



          So on one hand you have an alternating series and on the other hand you have something asymptotic to the harmonic series.






          share|cite|improve this answer





















          • Note that the convergence follows by the Leibniz criterion. :-)
            – p4sch
            Nov 30 '18 at 13:41














          1












          1








          1






          You have



          $$sinleft(frac{(n^2-1)pi}{n}right) = sinleft(npi -frac{pi}{n}right) $$



          $$=sin(npi)cos(pi/n) - cos(npi)sin(pi/n)$$



          $$= 0 +(-1)^{n+1}sin(pi/n).$$



          So on one hand you have an alternating series and on the other hand you have something asymptotic to the harmonic series.






          share|cite|improve this answer












          You have



          $$sinleft(frac{(n^2-1)pi}{n}right) = sinleft(npi -frac{pi}{n}right) $$



          $$=sin(npi)cos(pi/n) - cos(npi)sin(pi/n)$$



          $$= 0 +(-1)^{n+1}sin(pi/n).$$



          So on one hand you have an alternating series and on the other hand you have something asymptotic to the harmonic series.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 12:53









          B. GoddardB. Goddard

          18.4k21340




          18.4k21340












          • Note that the convergence follows by the Leibniz criterion. :-)
            – p4sch
            Nov 30 '18 at 13:41


















          • Note that the convergence follows by the Leibniz criterion. :-)
            – p4sch
            Nov 30 '18 at 13:41
















          Note that the convergence follows by the Leibniz criterion. :-)
          – p4sch
          Nov 30 '18 at 13:41




          Note that the convergence follows by the Leibniz criterion. :-)
          – p4sch
          Nov 30 '18 at 13:41


















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