Limit of an integral sum

First of all, lets find the recurrent relation for the $I_n$ . Integrating by parts we get $I_n=\int^1_0x^ne^x\,dx=e-\int^1_0nx^{n-1}e^x\,dx=e-nI_{n-1}.$ This is equivalent to $I_n+nI_{n-1}=e$ . The last equality implies $I_{n+1}+(n+1)I_n=I_n+nI_{n-1}$ or $I_{n+1}=n(I_{n-1}-I_n)$ . Now lets analyze our sum. We obtain $\sum_{k=1}^n\frac{I_{k+1}}{k}=\sum_{k=2}^n(I_{k-1}-I_k)+I_2=I_1-I_n+I_2.$

Now we must find to which number does $I_n$ tend. It's easy to notice that $0\leq \int^1_0x^ne^x\,dx \leq e\int^1_0x^n\,dx\Leftrightarrow 0\leq I_n\leq \frac{e}{n+1}.$

So we obtain that $\lim_{n\to\infty}I_n=0$

To finish our proof, we integrate by parts to calculate that $\lim_{n\to\infty} \left(I_1+I_2-I_n\right)=I_1+I_2=\boxed{e-1}.$

We know that : $\sum_{k=1}^{\infty} \frac{x^n}{n}=-\ln(1-x).$ Therefore : $L=\sum_{k=1}^{\infty} \frac{I_{k}}{k+1} =\int\limits_0^1 x e^x \sum_{k=1}^{\infty} \frac{x^k }{k} \ \mathrm{d}x$ $L.=- \int\limits_{0}^1 x e^x \ln(1-x)\ \mathrm{d}x.$ The interchange sum-integral is justified by the positiveness of all summands (monotone convergence theorem). $L=\bigg[-e^x ((x-1)\ln(x-1)+1\bigg]_0^1=1-e.$

$\displaystyle \int_0^1 x e^x \sum_{k \ge 1}\frac{x^k}{k}dx= -\int_0^1 x e^x \log(1-x)dx$

$\displaystyle = -\int_0^1 (1-t) e^{1-t} \log(t)dt=\int_0^1 t e^{1-t} \log(t)dt-\int_0^1 e^{1-t} \log(t)dt.$

Now, consider the integral

$\displaystyle I(n)=e \int_0^1 t^n e^{-t} dt \stackrel{\text{IBP}}{=}n!(e-1)-ne=\Gamma(n+1)(e-1)-ne$

$\displaystyle I'(n)=\Gamma'(n+1)(e-1)-e= e \int_0^1 t^n e^{-t} \log(t) dt.$

Thus our answer is equal to

$\displaystyle I'(1)-I'(0)=(\Gamma'(2)-\Gamma'(1))(e-1)=(1-\gamma+\gamma)(e-1)=e-1.$

We can modification $I_n$ as double Integral $I_n=\int_0^1 \! \int_0^1 x^{n+1}y^{n-1}e^x\,dx\,dy$

Since, $0<\int_0^1 \! \int_0^1 x^{2}e^x\,dx\,dy<1$ , so $I_n$ as geometric progression.

So, $\displaystyle \lim_{x \to \infty} \frac{I_{k+1}}{k}=\int_0^1 \! \int_0^1 \frac {x^2e^x}{1-xy}=\fbox{1.718}$

the limit can be written as{ - ∫x* e^x* log(1-x)dx }from x=0 to x=1