So I have the following series: $$\sum_{2n}^{4n} \frac 1k$$ and I need to find the limit of it when n appraoches infinity: $$\lim_{n\to\infty}\sum_{2n}^{4n} \frac 1k$$ I tried to consider this sum as the riemann sums of the function $$\frac1{x+1}$$ and thus calculating the integral of it (which seems right to me, but I don't know what are the boundaries of the integral itself, and also it's a pure guess which I don't know how to prove). Any kind of help would be great!

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    $\begingroup$ You're not far off; the difference is the Euler-Mascheroni constant. $\endgroup$
    – vadim123
    Apr 29 '15 at 20:08
  • $\begingroup$ The sequence is easily shown to converge to a number between $\;\frac12\;$ and $\;1\;$ . $\endgroup$
    – Timbuc
    Apr 29 '15 at 20:10
  • $\begingroup$ The function $\: k \mapsto \frac1k \:$ is monotone, so you should bound those sums between two integrals. $\hspace{.89 in}$ $\endgroup$
    – user57159
    Apr 29 '15 at 20:11
  • $\begingroup$ Hint: $~\displaystyle\sum_{2n}^{4n}\frac1k~=~H_{4n}-H_{2n-1} ~\approx~ \ln(4n)-\ln(2n-1)~=~\ln\frac{4n}{2n-1}.~$ Now take the limit as $n\to\infty$. $\endgroup$
    – Lucian
    Apr 29 '15 at 21:16

Assuming that the limit exists and is finite, we can compute it using Riemann sums since the function $f(x)=\frac{1}{x}$ is Riemann-integrable on $[2,4]$.

$$\lim_{n\to\infty}\sum_{k=2n}^{4n}\frac{1}{k}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=2n}^{4n}\frac{1}{\frac{k}{n}}\approx \int\limits_2^4\frac{\,\mathrm dx}{x}=\bigg[\ln|x|\bigg]_2^4=\ln4-\ln2=2\ln2-\ln2=\ln2$$

  • $\begingroup$ How did you choose the boundaries to be 4 and 2? No doubt it makes sense since the lower boundary of the sum is 2n and the upper is 4n, but how do I justify it mathematically? $\endgroup$
    – Calculus
    Apr 29 '15 at 20:29
  • $\begingroup$ @Calculus, By the Riemann sum argument of a definite integral, we have, for a Riemann-integrable function $f(x)$ on $[a,b]$, $nh=b-a$ such that $n\to\infty$ and $h\to 0$. And, $$\lim_{h\to 0}h\sum_{k=0}^{n-1}f(a+kh)=\lim_{h\to 0}h\sum_{k=1}^nf(a+kh)=\int\limits_a^bf(x)\,\mathrm dx$$ $\endgroup$ Apr 29 '15 at 20:35
  • $\begingroup$ Can you work it out yourself using this hint? $\endgroup$ Apr 29 '15 at 20:35

Using the integral test idea, $$ \int_0^{N-1} \frac{dx}{x+1} \leqslant \sum_{k=1}^N \frac{1}{k} \leqslant \int_0^{N} \frac{dx}{x+1}. $$ The integrals evaluate to $\log{N}$ and $\log{(N+1)}$. Now you can take $N=4n$, subtract the version of this with $N=2n$, and you find $$ \log{4n}-\log{(2n-1)} \leqslant \sum_{k=1}^N \frac{1}{k} \leqslant \log{(4n+1)}-\log{(2n)}. $$ You can then show that the limit as $n \to \infty$ on both outer sides of this inequality is $\log{2}$, so the limit of the sum must also be $\log{2}$ by the squeeze theorem.


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