Find out the sum of the series $\displaystyle \sum\limits_{n=1}^{\infty} \dfrac{n^2}{ 2^n}$. I have checked the convergence, but how to calculate the sum?
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$\begingroup$ See polylogarithm. $\endgroup$– LucianJan 19, 2014 at 0:51
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2 Answers
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Note $\sum\limits_{n\geqslant 0} t^n = \dfrac{1}{ 1-t}$ gives $\sum\limits_{n\geqslant 1} nt^n = \dfrac{t}{( 1 - t)^2}$ after differentiation and multiplication by $t$, which in turn gives $\sum\limits_{n\geqslant 1} n^2t^n = \dfrac{t(t+1)}{( 1 - t)^3}$ by the same token.
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Begin with a geometric series, $$\sum_{n=0}^\infty z^n = {1\over 1 - z}.$$ if $|z| < 1$. Differentiate and you get $$\sum_{n=1}^\infty nz^{n-1} = {1\over (1-z)^2}.$$ Do it again $$\sum_{n=2}^\infty n(n-1)z^{n-2} = {1\over (1-z)^3}.$$ Massage these, plug in $z = 1/2$ and you will get it.
answered Jan 19, 2014 at 0:47