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Let $T_n=\sum _{k=1}^{n}\dfrac{e^{k-1}}{\pi ^{k+1}}$ calculate the $\lim_{n\to\infty}T_n$

Note $T_n$ is a geometric series:

\begin{align*} T_n&=\sum _{k=1}^{n \:}\dfrac{e^{k-1}}{\pi ^{k+1}}\\ &= \pi^{-2}.\sum _{k=1}^{n \:}\dfrac{e^{k-1}}{\pi ^{k-1}}\\ &= \pi^{-2}.\sum _{k=1}^{n \:}\left(\dfrac{e}{\pi }\right)^{k-1}\\ &= \pi^{-2}.\sum _{k=0}^{n-1 \:}\left(\dfrac{e}{\pi }\right)^{k}(\text{change of index})\\ &= \pi^{-2}.\dfrac{1-\left(\dfrac{e}{\pi }\right)^{n}}{1-\left(\dfrac{e}{\pi }\right)}(\text{Geometric series})\\ &= \pi^{-1-n}.\dfrac{\pi^n-e^n }{\pi-e }(\text{Geometric series})\\ \text{then the limit of $T_n$ }\\ \lim_{n\to\infty}T_n&=\lim_{n\to\infty} \pi^{-1-n}.\dfrac{\pi^n-e^n }{\pi-e } ?? \end{align*} I'm stuck in limit.

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    $\begingroup$ Restart at the line: $\pi^{-2} \frac{1-(e/\pi)^n}{1-(e/\pi)}$. Now observe that $(e/\pi) < 1$ so $(e/\pi)^n \to 0$. $\endgroup$
    – Winther
    Commented Jul 30, 2014 at 23:44

3 Answers 3

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You just need to take the limit one step earlier. That is, find

$$\lim_{n\to\infty}\pi^{-2}\frac{1-\left(\dfrac{e}{\pi}\right)^n}{1-\dfrac{e}{\pi}}$$

by using the fact that if $|x| < 1$ then $\displaystyle\lim_{n \to \infty}x^n = 0$.

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At the line where you cite "change of index" you have $T_{n}$ in the form you need to apply the convergence formula for geometric series, $\sum_0^\infty ar^{n}= \frac{a}{1-r}$. You found $a=\pi^{-2}$ and $r=\frac{e}{\pi}$ already by correctly rewriting the infinite series, so it is just a matter of plugging the relevant numbers in.

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Just to make the problem more general, let us consider $$T_n=\sum _{k=c}^{n \:}\dfrac{x^{k+a}}{y^{k+b}}=\dfrac{x^{a}}{y^{b}}\sum _{k=c}^{n \:}z^k$$ where $z=\frac{x}{y}$. $$T_n=\dfrac{x^{a}}{y^{b}}\frac{z^{n+1}-z^c}{z-1}$$ In order to have a defined limit value when $n$ goes to infinity, as for any geometric series, it is required that $z<1$. So, under this condition, $$\lim_{n\to\infty}T_n=\dfrac{x^{a}}{y^{b}}\frac{z^c}{1-z}$$

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