Proving convergence of sum to 2n

Question

I need to prove that the following sequence converges as $n\to\infty$: $\sum\limits_{k=n+1}^{2n} 1/k$

The problem is that I've only ever seen sums from i to n for example. I'm confused because not only is 2n alien to me, but also k and n are related.

So far I've tried substituting k for n+1 to simplify the expression, but it led nowhere. I also tried to enter this into spreadsheet calculation in hopes to get an idea where the border might be, but the result just grows and grows. I almost expected this because the sequence is reminiscent of the harmonic series which diverges. The key difference is obviously the beginning and end of the sum, but I can't figure out how to tackle this.

Any ideas or pointers are appreciated

You're summing a finite number of terms, hence your final sum will be finite and it will converge. — EA304GT, Nov 21 '15 at 19:21
@EA304GT: I'm afraid things are not so simple. To take an obvious counter-example, if the summation term was $1$ instead of $1/k$, the sequence would be $(1,2,3,\ldots)$, which certainly doesn't converge. — TonyK, Nov 21 '15 at 19:23
Does this answer your question? The limit of truncated sums of harmonic series, $\lim\limits_{k\to\infty}\sum_{n=k+1}^{2k}{\frac{1}{n}}$ — Arctic Char, Sep 30 '20 at 05:21

score 3 · Accepted Answer · answered Nov 21 '15 at 19:24

3

Observe that you sum $n$ terms in the sum, hence you can estimate the sum by $n$ times the biggest term, hence you get $$ \sum_{k=n+1}^{2n}\frac{1}{k}\leqslant \frac{n}{n+1}\stackrel{n\to\infty}{\longrightarrow}1 $$ Hence the sequence is bounded and since it is monotonically increasing the convergence follows.

answered Nov 21 '15 at 19:24

frog

2,381

1

The assertion that the sequence is monotonically increasing could use a bit of justification. – Eric Wofsey Nov 21 '15 at 19:33
So I don't need to find out how it is bounded? I can simply claim it is, because there is a greater example? – John Nov 21 '15 at 19:33
@Eric: Yes, I'm sorry for being a bit sloppy but it is explained now in another answer. – frog Nov 21 '15 at 19:34
@John: it is clearly bounded by $1$! – Bernard Nov 21 '15 at 19:43

score 0 · Answer 2 · answered Nov 21 '15 at 19:31

Let us write $a_n=\sum_{k=n+1}^{2n} 1/k$, and let's compare $a_n$ to $a_{n+1}$. They have almost the same terms, except $a_n$ has one more term at the beginning and $a_{n+1}$ has two more terms at the end. We get that $$a_{n+1}-a_n=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}>\frac{1}{2n+2}+\frac{1}{2n+2}-\frac{1}{n+1}=0.$$

Thus the numbers $a_n$ are increasing. Now note that for each $n$, every term in the sum for $a_n$ is at most $1/(n+1)$, and there are $n$ terms, so $a_n\leq n/(n+1)<1$. So the sequence $a_n$ is monotone increasing and bounded above by $1$. It follows that it must converge.

Proving convergence of sum to 2n

2 Answers2