Calculate $\sum_{n=1}^{\infty} (V(n)-L)$ Since I have $$V(n)=(n^2+1)\int_{0}^{\pi} e^{-nx}\sin x dx$$
And I calculated $V(1)=e^{-\pi}-1$ and $L=\lim_{n\to\infty} V(n)=1$
And how can I calculate $\sum_{n=1}^{\infty} (V(n)-L)$
Thank in advanced!
 A: By integrating by parts twice, you can easily get that $$\frac{V(n)}{n^2+1}=\int_0^π e^{-n x} \sin(x) dx = \frac{e^{-n\pi} + 1}{n^2 + 1}$$
Then since $V(n)\rightarrow 1$, you're just looking to $$\sum_{k=1}^{+\infty} e^{-k\pi} = \sum_{k=1}^{+\infty} (e^{-\pi})^k = \frac{e^{-\pi}}{1-e^{-\pi}}$$
A: You can generalize above formula for all $n$ using the product rule for integration twice. Take $u'=e^{-nx}$ and $v=\sin x$ for the first application and $u'=e^{-nx}$ and $v=\cos x$ for the second application of this product rule:
\begin{align}
\int_0^\pi e^{-nx}\sin x dx &= \left[-\frac 1 n e^{-nx} \sin x\right]_0^\pi +\frac 1 n \int_0^\pi e^{-nx} \cos xdx \\
&=0 +\frac 1 n \left(\left[-\frac 1 n e^{-nx} \cos x\right]_0^\pi- \frac 1 n \int_0^\pi e^{-nx} \sin xdx\right) \\
&=\frac 1 n \left( \frac{1}{n}e^{-n\pi} +\frac 1 n\right)- \frac{1}{n^2} \int_0^\pi e^{-nx} \sin xdx \\
&= \frac{1}{n^2} e^{-n\pi}+\frac{1}{n^2} - \frac{1}{n^2}\int_0^\pi e^{-nx}\sin x dx
\end{align}
This equation can now be reformulated to get finally:
\begin{align}
\int_0^\pi e^{-nx}\sin x dx &= \frac{1}{n^2} e^{-n\pi}+\frac{1}{n^2} - \frac{1}{n^2}\int_0^\pi e^{-nx}\sin x dx \\
\Rightarrow \left(1+\frac{1}{n^2}\right)\int_0^\pi e^{-nx}\sin x dx &= \frac{1}{n^2} (e^{-n\pi}+1) \\
\Rightarrow \int_0^\pi e^{-nx}\sin x dx &= \frac{1}{(n^2+1)} (e^{-n\pi}+1)
\end{align}
Concluding everything you get
$$V(n) = e^{-n\pi} + 1$$
Now you can enter everything into the infinite sum to get
\begin{align}
\sum_{n=1}^\infty ((e^{-n\pi} + 1) - 1) = \sum_{n=1}^\infty e^{-n\pi} = \sum_{n=1}^\infty (e^{-\pi})^n= \sum_{n=0}^\infty (e^{-\pi})^n - 1
\end{align}
As $e^{-\pi}< 1$ you have a geometric series and thus the final result is
$$
\sum_{n=1}^\infty (V(n)-L) = \frac{1}{1-e^{-\pi}} - 1 = \frac{e^{-\pi}}{1-e^{-\pi}} =\frac{1}{e^{\pi}-1}\approx 0.045
$$
A: $$\begin{align}V(n)=&(n^2+1)\quad\int_0^\pi e^{-nx}\sin(x)\,\mathrm dx\\
=&(n^2+1)\quad\Im\int_0^\pi e^{(j-n)x}\,\mathrm dx\\
=&(n^2+1)\quad\Im\left[\frac{e^{(j-n)x}}{j-n}\right]_0^\pi\\
=&(n^2+1)\quad\Im\left\{\frac{(j+n)\left(e^{(j-n)\pi}-1\right)}{-(n^2+1)}\right\}\\
=&\Im\left\{(j+n)\left(e^{-n\pi}+1\right)\right\}\\
=&e^{-n\pi}+1\end{align}\tag{V}$$

$$\begin{align}L=&\lim_{n\to\infty}V(n)\\
=&\lim_{n\to\infty}\left(e^{-n\pi}+1\right)\\
=&\,1\end{align}\tag{L}$$

$$\begin{align}S=&\sum_{n=1}^\infty\left[V(n)-L\right]\\
=&\sum_{n=1}^\infty e^{-n\pi}\\
=&\sum_{n=1}^\infty\left(e^{-\pi}\right)^n\\
=&\sum_{n=0}^\infty\left(e^{-\pi}\right)^n\,\,-\left.e^{-n\pi}\right|_{n=0}\\
=&\sum_{n=0}^\infty\left(e^{-\pi}\right)^n-1\qquad(\text{geometric series})\\
=&\frac{1}{1-e^{-\pi}}-1\\
=&\frac{e^{-\pi}}{1-e^{-\pi}}\color{red}{\times\frac{e^\pi}{e^\pi}}\\
=&\frac{1}{e^{\pi}-1}\end{align}\tag{S}$$
