TITLE: If $I_{n} = \int_{0}^{1}x^n\cdot e^xdx,$ Then $\lim_{n\rightarrow \infty}\left(\sum_{k=1}^{n}\frac{I_{k+1}}{k}\right) =$
QUESTION [5 upvotes]: (1) If $\displaystyle I_{n} = \int_{0}^{1}x^n\cdot e^xdx$. Then $\displaystyle \lim_{n\rightarrow \infty}\left(\sum_{k=1}^{n}\frac{I_{k+1}}{k}\right) = $
$(2)$ Value of $\displaystyle \frac{\int_{0}^{\frac{\pi}{2}}e^{-x}\sin^{n}(x)dx}{\int_{0}^{\frac{\pi}{2}}e^{x}\sin^{n}(x)dx} = \;, $ Where $n\in \mathbb{N}$
$\bf{My\; Try}::$ for $(1)$ one:: Given $\displaystyle I_{n} = \int_{0}^{1}x^n\cdot e^xdx$
$\displaystyle \Rightarrow I_{n} = \left[x^n\cdot e^x\right]_{0}^{1}-n\int_{0}^{1}x^{n-1}\cdot e^xdx = e-nI_{n-1}\Rightarrow I_{n} = e-nI_{n-1}$
$\Rightarrow I_{n}+n\cdot I_{n-1} = e$
Now I did not understand how can i solve after that
Help Required.
Thanks
$(2)$ Let $\displaystyle I_{n} = \int_{0}^{\frac{\pi}{2}}e^{-x}\sin^{n}(x)dx$ and $\displaystyle J_{n} = \int_{0}^{\frac{\pi}{2}}e^{x}\sin^{n}(x)dx$
Now I did not understand how can i solve after that
Help Required
Thanks

REPLY [4 votes]: Because $x \in (0,1)$, the sum
$$\sum_{k=1}^n \frac{x^k}{k} $$
converges as $n \to \infty$.  Thus we can rearrange the integration and summation:
$$\sum_{k=1}^{\infty} \frac{I_{k+1}}{k} = \int_0^1 dx \, x \,e^x \sum_{k=1}^{\infty} \frac{x^k}{k} = -\int_0^1 dx \, x \,e^x \, \log{(1-x)} $$
The integral is easily evaluated using integration by parts:
$$-\int_0^1 dx \, x \,e^x \, \log{(1-x)} = \left [(1-x) e^x \log{(1-x)} \right ]_0^1+ \int_0^1 dx \frac{1-x}{1-x} e^x$$
which is equal to $e-1$.  Thus, the limit is equal to $e-1$.