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I am having trouble trying to figure out the solution to the infinite series $ \sum_{i=1}^{\infty} \left( \frac{1}{2} \right)^i \left( \frac{1}{i}\right)$.

What kind of infinite series is this, and what would be the best approach to simplifying? My intuition tells me it's a converging series, but I'm lost on how to approach this problem.

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    $\begingroup$ Do you know the power series for $\log(1-z)$? $\endgroup$ – metamorphy Jul 24 '20 at 23:07
  • $\begingroup$ We have $\sum_{i=1}^\infty\frac{x^i}{i}=-\ln(1-x) \Longrightarrow \sum_{i=1}^\infty\frac{(1/2)^i}{i}=-\ln(1-1/2)=\ln(2).$ $\endgroup$ – Ali Shadhar Jul 25 '20 at 7:23
  • $\begingroup$ @metamorphy How was this series derived? I don't have much familiarity with power series, but with taylor series expansion there's a generic formula. Is there an analogous formula for power series expansions? $\endgroup$ – David Jul 25 '20 at 18:42
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Note that $\frac{a^k}{k} = \int_{0}^{a}x^{k-1}dx$, the interchange the sum and integral, get the (convergent) limit of the sum, and solve the integral for $x$.

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    $\begingroup$ I feel this sort of post would be more suited for a comment, unless you would like to flesh out your work. $\endgroup$ – Ninad Munshi Jul 24 '20 at 23:11
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Let's get some terms!

We find $\frac{1}{2}$, $\frac{1}{8}$, $\frac{1}{24}$, $\frac{1}{64}$, $\frac{1}{160}, $\frac{1}{384}$.

These don't look too good. I kept finding some terms and added them up, it looks to converge around $\frac{7}{10}$.

Wolfram alpha says 0.693147180559945309417232121458176568075500134360255254120...

Hope this helped!

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