Does $\displaystyle\sum_{k=1}^{\infty} \left (\sum_{j=1}^{k}\frac 1 j\right)^{-k}$ converges ?

Let's call the inner sum $a_k$ such that $\displaystyle\sum_{k=1}^{\infty} (a_k)^{-k}$, applying root test we get: $(a_k)^{-1}= \left(\sum_{j=1}^{k}\frac 1 j\right)^{-1} = \frac {1}{1/1+1/2+...1/k}<1$

So the given sum converges. Is that all right ?

  • 1
    $\begingroup$ Well, one wants to take the limit of the $|(a_k)^{-1}|$. But as the harmonic series diverges, this limit is then $0$ - so the sum converges. $\endgroup$ – user98602 Feb 26 '14 at 11:55
  • $\begingroup$ It would be interesting to see the $(k+1)^{th}$ term expressed in terms of the $k^{th}$ one. $\endgroup$ – Nikolaj-K Feb 26 '14 at 12:06
  • $\begingroup$ @NikolajK. what do you mean ? $\endgroup$ – GinKin Feb 26 '14 at 12:23
  • $\begingroup$ @GinKin: Go to WolframAlpha.com and type Sum[Sum[1/j,{j,1,k}]^(-k),{k,1,Infinity}], it converges to $1.68227..$. Can we actually compute that? Thoughts: Every partial sum here is some rational number and the frist three terms in $k$ sum to $\frac{19247}{11979}=1.6067..$. Further, note that $\sum_{j=1}^4\frac{1}{j}=1+1/2+1/3+1/4>2$. So your sum equals $\frac{19247}{11979}+\sum_{k=4}^\infty \frac{1}{\left(2+\varepsilon_k\right)^k}$ with $\varepsilon_k$ some positive numbers. Even for $\varepsilon_k=0$ the remaining sum would only be $\frac{1}{8}$. I tried to compute the consecutive grow. $\endgroup$ – Nikolaj-K Feb 26 '14 at 12:46

A nice idea in case you encounter the harmonic series again:

notice that $\sum_{k=1}^{\infty} \left (\sum_{j=1}^{k}\frac 1 j\right)^{-k}$ is convergent, as is $ \sum_{k=2}^{\infty} \frac{1}{( \ln k)^k}$ because $\sum_{j=1}^{k}\frac 1 j \ge \ln k$

Now since $\ln k >1$ for $k>3$ the series obviously converge by comparison with the geometric series with general term $\frac{1}{ (\ln 4)^n}$.

To answer your comment:

In a sense, yes. First check that

$$\frac{1}{k}=\int_{k}^{k+1}\sup_{x \in [k,k+1]}(\frac{1}{x})dt \ge \int_{k}^{k+1} \frac{dt}{t} \ge \int_{k}^{k+1}\inf_{x \in [k,k+1]}(\frac{1}{x})dt = \frac{1}{k+1}.$$

Then calculating the integral in the middle you get $$\int_{k}^{k+1} \frac{dt}{t}=\ln(k+1) - \ln(k) .$$

Summing things up to $N$ we obtain: $$\sum_1^N \frac{1}{k} \ge \sum_{1}^N(\ln(k+1)-\ln(k))\ge \sum_1^N \frac{1}{k+1}.$$ To finish things, notice that you can telescope the expression in the middle and get $\ln(1+N) - \ln(1)=\ln(N+1)$.

  • $\begingroup$ @AntonioVargas You are right, i should have been more careful. Is it ok now? $\endgroup$ – Marko Karbevski Aug 1 '14 at 0:17
  • $\begingroup$ I think some clarification of "is of the same nature" would help, but otherwise it looks good. $\endgroup$ – Antonio Vargas Aug 1 '14 at 0:21
  • $\begingroup$ what are using to justify: $\sum_{j=1}^{k}\frac 1 j \sim \ln k$ ? is it because of the derivative of $\ln k$? I'm trying to understand the legitimacy of this move so I could use it properly $\endgroup$ – Jneven Jul 3 '18 at 7:51
  • $\begingroup$ @Jneven Please see the edit. If anything is not fully clear, feel free to ask again. $\endgroup$ – Marko Karbevski Jul 3 '18 at 14:11
  • $\begingroup$ oh okay. I am taking the class of "calculus 1a". we got this question as a HW, but integrals is not a part of the metrical (it's in the class "calculus 2a") , so i'm wondering how to solve it without using integrals. $\endgroup$ – Jneven Jul 3 '18 at 15:44

Starting with the second term, the $k$th term is at most $(1 + 1/2)^{-k} = (2/3)^k$, so since the geometric series $\sum_{k=2}^{\infty} (2/3)^k$ converges so does the original series.


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