Autor Tema: Integral and summation

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09 Mayo, 2019, 12:30 pm
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jacks

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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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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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