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$$\sum_{k=1}^{∞}\frac{1}{(2^k-1)}$$

I need to show that the series is convergent.

If there is a method that can be used to solve this (not the comparison test), what test would that be, and how would I go about solving it using that test?

gticecream8
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  • The comparison test it relatively difficult to use it you want to find the sum of a series. On the other hand, it can be very effective if you want to prove that a given series converges, without caring much about what it converges to. What kind of solution are you looking for? – Arthur Nov 09 '16 at 13:32
  • @Arthur sorry about that, edited to clarify–looking to show that it converges – gticecream8 Nov 09 '16 at 13:33

5 Answers5

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Compare the general element of the series with $\frac{1}{2^k}$, and since the series $\displaystyle\sum_{k=1}^{\infty} \frac{1}{2^k}$ converges, we have the convergence of the first series by using the (second) complarison test.

By compare I mean evaluate the limit $\displaystyle\lim_{k\rightarrow \infty} \frac{\frac{1}{2^k-1}}{\frac{1}{2^k}}.$

MKBG
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  • Thanks–I realized this soon after I asked the question, and edited the question, but this answers my initial one perfectly. – gticecream8 Nov 09 '16 at 13:34
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You can use the Root Test: $$\lim_{k\to +\infty}\frac{1}{(2^k-1)^{1/k}}= \lim_{k\to +\infty}\frac{1}{2(1-2^{-k})^{1/k}}=\frac{1}{2\cdot 1^{0}}=\frac{1}{2}<1.$$ which implies that the series is convergent.

Robert Z
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  • While you definitely can use the root test, you might want to explain how the denominator approaches $2$ a bit more formally, as the OP may not be able to do so. Thinking to what I would know were I the OP, I believe I would jump on the Binomial Expansion at this point to try to show that $2^k$ dominates. Another option is to take logarithms and use L'Hopital's rule. Regardless of whether you add something along these lines, I thought it important the OP knows how to. – Brevan Ellefsen Nov 09 '16 at 14:00
  • @Brevan Ellefsen Is it better now? – Robert Z Nov 09 '16 at 14:07
  • yes, I definitely think so. +1 – Brevan Ellefsen Nov 09 '16 at 14:11
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The Ratio test! This tells us that for a series $\sum\limits_{k=1}^{\infty}a_k$, we should evaluate the limit $$\lim\limits_{k \to \infty}\left|\frac{a_{k+1}}{a_k}\right| = L$$ If $L > 1$ the series diverges and if $L < 1$ the series converges (absolutely). If $L = 1$ or no limit exists, no conclusion can be made.

In your case, we consider the limit $$\lim_{k \to \infty} \frac{\frac{1}{2^{k+1}-1}}{\frac{1}{2^k-1}} = \lim_{k \to \infty} \frac{2^k-1}{2^{k+1}-1} = \lim_{k \to \infty} \frac{1-\frac{1}{2^k}}{2-\frac{1}{2^k}} = \frac{1}{2}$$

Since $\frac{1}{2} < 1$, the Ratio test tell us the series converges.

Community
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sTertooy
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  • The case $L=1$ doesn't mean no conclusion can be made but is instead inconclusive. There is a difference, as one can expand the test a bit to be able to handle the case $L=1$. The information is still there, you just have to do a little more! – Brevan Ellefsen Nov 09 '16 at 13:54
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Just as an alternate technique, the convergence of the series isn't too hard to show with the "Ratio Test". Remembering that we need $\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right|< 1$ to show the series converges absolutely, we find the following: $$\lim_{k\to\infty} \frac{2^k-1}{2^{k+1}-1}=\lim_{k\to\infty} \frac{1-2^{-k}}{2-2^{-k}}= \frac 12 < 1$$

Brevan Ellefsen
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the sum it is given by Mathematica is given by $$\frac{\log (2)-\psi _{\frac{1}{2}}^{(0)}(1)}{\log (2)}$$ and containes the polygamma function.

Dr. Sonnhard Graubner
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