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December 30th, 2018, 04:05 AM   #1
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Series Sum

Evaluation of $$\sum^{\infty}_{k=1}\frac{1}{k\cdot 2^k}$$
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December 30th, 2018, 04:46 AM   #2
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Take the derivative of $\sum_k \frac{x^k}{k}$
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January 2nd, 2019, 09:55 AM   #3
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K --> infinite
2^K--> infinite
K*2^K --> infinite

Therefore, the result is 0
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January 2nd, 2019, 10:59 AM   #4
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Yes, the terms must converge to zero for the series to converge (and they do, as you point out). No, the series doesn't converge to zero.
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January 2nd, 2019, 11:19 AM   #5
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If not, how can we evaluate it?
I know the method of derivatives, but it is not working.

Last edited by skipjack; January 2nd, 2019 at 04:52 PM.
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January 2nd, 2019, 11:53 AM   #6
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Quote:
Originally Posted by oldbrother View Post
K --> infinite
2^K--> infinite
K*2^K --> infinite

Therefore, the result is 0
That only means that the larger k is the smaller $\displaystyle \dfrac{x^k}{k}$ becomes. It says nothing about the sum.

-Dan
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January 2nd, 2019, 01:53 PM   #7
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$\displaystyle f(x)=\sum_{k=1}^\infty \frac{x^k}{k},\ f'(x)=\sum_{k=1}^\infty x^{k-1}=\frac{1}{1-x}$. Therefore $f(x)=-\ln(1-x)$ or $f(\frac{1}{2})=\ln(2)$.
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Last edited by skipjack; January 2nd, 2019 at 04:41 PM.
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