### Autor Tema: Integral and summation

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09 Mayo, 2019, 12:30 pm
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#### jacks

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##### Integral and summation
If $\displaystyle I_{k}=\int^{2}_{1}e^{x-1}x^{-n}dx.$ Then value of $\displaystyle \sum^{\infty}_{k=0}\bigg(1-kI_{k}\bigg)$ is

11 Enero, 2020, 02:42 am
Respuesta #1

#### kike0001

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##### Re: Integral and summation
Hello I assumed that
$\displaystyle I_{k}=\int^{2}_{1}e^{x-1}x^{-k}dx.$
Then
$\displaystyle I_{k-1}=\int^{2}_{1}e^{x-1}x^{-k+1}dx$
by parts, with $u=x^{-k+1}$ and $dv=e^{x-1}dx$:
$\displaystyle I_{k-1}=2^{-k+1}e-1+k\int_1^2e^{x-1}x^{-k}dx-\int_1^2e^{x-1}x^{-k}dx$
$kI_k=I_{k-1}+I_k+1-2^{-k+1}e$

then:
$1-kI_k=2^{-k+1}e-I_{k-1}-I_k$

$\displaystyle \sum^{\infty}_{k=0}\bigg(1-kI_{k}\bigg)=\displaystyle \sum^{\infty}_{k=0}\bigg(2^{-k+1}e-(I_{k-1}+I_k)\bigg)$

If $a_k=2^{-k+1}e$ and $b_k=-(I_{k-1}+I_k)$

note that $\displaystyle \sum^{\infty}_{k=0}a_k=\displaystyle \sum^{\infty}_{k=0}\bigg(2^{-k+1}e\bigg)=4e$ this series converges

now
$\displaystyle \sum^{\infty}_{k=0}b_k=\displaystyle \sum^{\infty}_{k=0}\bigg[-(I_{k-1}+I_k)\bigg]=\displaystyle \sum^{\infty}_{k=0}\bigg[-\int^{2}_{1}e^{x-1}x^{-k}(x+1)dx\bigg]$

$\displaystyle \sum^{\infty}_{k=0}\bigg[-\int^{2}_{1}e^{x-1}x^{-k}(x+1)dx\bigg]=-\int^{2}_{1}e^{x-1}(x+1)\sum^{\infty}_{k=0}(x^{-k})dx=-\int^{2}_{1}e^{x-1}(x+1)\frac{x}{x-1}dx$

How $\frac{x}{x-1}\leq e^{x-1}(x+1)\frac{x}{x-1}$ and $\int^{2}_{1}\frac{x}{x-1}dx$ diverges then $\int^{2}_{1}e^{x-1}(x+1)\frac{x}{x-1}dx$ also diverges

then $\displaystyle \sum^{\infty}_{k=0}b_k$ diverges

therefore $\displaystyle \sum^{\infty}_{k=0}\bigg(a_k+b_k\bigg)=\sum^{\infty}_{k=0}\bigg(1-kI_k\bigg)$ diverges

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20 Enero, 2020, 10:55 am
Respuesta #2

#### jacks

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Thanks kike001