Thursday, January 3, 2019

real analysis - $sum limits_{n=1}^{infty}n(frac{2}{3})^n$ Evalute Sum











How can you compute the limit of
$\sum \limits_{n=1}^{\infty} n(2/3)^n$



Evidently it is equal to 6 by wolfram alpha but how could you compute such a sum analytically?


Answer



$$
\begin{align*}

\sum_{n=1}^\infty n(2/3)^n &=
\sum_{m=1}^\infty \sum_{n=m}^\infty (2/3)^n \\ &=
\sum_{m=1}^\infty \frac{(2/3)^m}{1-2/3} \\ &=
\frac{2/3}{(1-2/3)^2} = 6.
\end{align*}
$$


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