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I have to show whether $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ is convergent or divergent. I tried using the squeeze theorem to prove it was convergent. So what I did was bound ${y_n}$ in between $\frac{-1}{n}$ and $\frac{1}{n}$. That is $\frac{-1}{n} \leq \frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n} \leq \frac{1}{n}$ Since we also proved earlier that $lim\frac{1}{n}=0$ and $lim\frac{-1}{n}= -lim\frac{-1}{n}=0$ It follows by the squeeze theorem that $lim(y_n)=0$. Would this be correct?

Edit: This what I have now. Well this what I get so far that $ \sum_{k=n+1}^{2n}\frac{1}{k}≥ \sum_{k=n+1}^{2n}\frac{1}{k+1}$ Hence its divergent. I think it works since $ \sum_{k=n+1}^{2n}\frac{1}{k+1}=\frac{1}{(n+1)+1}+\frac{1}{(n+2)+1}+...+\frac{1}{2n+1}$ ≤ $ \sum_{k=n+1}^{2n}\frac{1}{k}= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} $.

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I think the right half of your squeeze is wrong. For $n=2$, it gives $(1/3)+(1/4)\le1/2$. – Gerry Myerson Jul 6 '13 at 5:39
oh I did prove that $lim(\frac{b}{n})=0$ where $b\in\mathbb{R}$. I think I can try using that. – user60887 Jul 6 '13 at 5:41
oh ok so its divergent what if I index n such that $y_n=\frac{1}{n_1+1}+\frac{1}{n_2+2}+..+\frac{1}{n_k^2}$ such that $n=n_1=...=n_k$ Then it follows that $y_n=\frac{1}{n_1+1}+\frac{1}{n_2+2}+..+\frac{1}{n_k^2}$ < $\frac{1}{n_1+1}+\frac{1}{n_2+1}+..+\frac{1}{n_k+1}$. Hence it is divergent. – user60887 Jul 6 '13 at 5:53
Sorry, I don't follow you. – Gerry Myerson Jul 6 '13 at 5:56
I think that the recent change of the question ($n^2$ changed to $2n$) is inappropriate. The old version was already answered, and it has been removed. It would be better to ask the new question as a new question, in a separate thread (with a link to this one). (Even though apparently the original version was posted by mistake, it became its own question.) – Jonas Meyer Jul 7 '13 at 1:34
up vote 1 down vote accepted

If you know that

$$\gamma_n=\frac{1}{1}+\frac{1}{2}+..+\frac{1}{n}- \ln(n) \,,$$ is convergent then your sequence is exactly

$$\gamma_{n^2}-\gamma_n+ \ln(n) \,.$$

As $\gamma_{n^2}-\gamma_n \to 0$ and $\ln(n) \to \infty$ it follows that your sequence diverges to $\infty$.

Edit With the change, your sequence is

$$\gamma_{2n}-\gamma_n+\ln(2n)-\ln(n)= \gamma_{2n}-\gamma_n+\ln(2)$$

which is convergent.

A simpler solution your sequence is also

$$\frac{1}{n} \sum_{k=1}^n \frac{1}{1+\frac{k}{n}}$$

which is the Riemann Sum associated to $f(x)=\frac{1}{x}$ on $[1,2]$ with $x_k=x_k^*=1+\frac{k}{n}$.

Both solutions also Yield $\ln(2)$ as the limit.

Third solution


is a well known identity, pretty standard induction problem. Then you can use the Alternating Series Test.

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I have made a bad mistake in the problem. The last term $\frac{1}{n^2}$ is supposed to be $\frac{1}{2n}$ – user60887 Jul 7 '13 at 0:54
@user60887 Gave new solutions to the new problem... – N. S. Jul 7 '13 at 1:30

Your reasoning would be correct if $y_n<\frac{1}{n}$, which is not true. The sequence actually diverges. To prove it, we use the fact that for $k\in[\![1,n-1]\!]$, \begin{align} \frac{1}{kn+1}+\frac{1}{kn+2}+...+\frac{1}{kn+n} &\geq \frac{1}{kn+n}+\frac{1}{kn+n}+...+\frac{1}{kn+n} \\ &\geq \frac{n}{n(k+1)}=\frac{1}{k+1} \end{align}

Then, we can rewrite $y_n$ using such partial sums: \begin{align} y_n&=\sum_{k=1}^{n-1} \frac{1}{kn+1}+\frac{1}{kn+2}+...+\frac{1}{kn+n}\\ &\geq\sum_{k=1}^{n-1}\frac{1}{k+1} \end{align} This sum diverges when $n\to\infty$, thus $y_n$ also diverges.

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Note that $\frac1{k+1} \le \int_k^{k+1} \frac{dt}{t} \le \frac1{k} $.

Summing from $n+1$ to $n^2$, $\sum_{k=n}^{n^2-1}\frac1{k+1} \le \sum_{k=n}^{n^2-1}\int_k^{k+1} \frac{dt}{t} \le \sum_{k=n}^{n^2-1}\frac1{k} $, or $\sum_{k=n+1}^{n^2}\frac1{k} \le \int_n^{n^2} \frac{dt}{t} \le \sum_{k=n}^{n^2-1}\frac1{k} $.

Using the right-hand inequality, $ \sum_{k=n+1}^{n^2}\frac1{k} =-\frac1{n}+\frac1{n^2}+\sum_{k=n}^{n^2-1}\frac1{k} \ge -\frac1{n}+\frac1{n^2}+\int_n^{n^2} \frac{dt}{t} >-\frac1{n}+ \ln(n^2)-\ln(n) = \ln(n)-\frac1{n} $ which diverges as $n \to \infty$.

If we use the left-hand inequality, $ \sum_{k=n+1}^{n^2}\frac1{k} \le \ln(n) $.

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Here is how to proceed using the Riemann sum idea

$$ y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}=\sum_{i=1}^{n}\frac{1}{n+i}$$

$$=\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+\frac{i}{n}} \longrightarrow_{n\to \infty} \int_{0}^{1}\frac{dx}{1+x}=\dots.$$

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The sum $$y_n:=\sum_{k=n+1}^{k=n^2}\frac 1 k = \Psi(n^2)-\Psi(n+1),$$ where $\Psi(x):= \frac {\Gamma'(x)} {\Gamma(x)} \sim \ln \left( x \right) -\frac 1 2\,{x}^{-1}-\frac1 {12}\,{x}^{-2}+O \left( {x}^{-4} \right) $ as $x \to \infty $. In view of this $y_n \to \infty$ as $n \to \infty.$

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True, but unlikely to help OP any time in the next two or three years. – Gerry Myerson Jul 6 '13 at 6:08

If $y_n=\sum\limits_{k=n+1}^{2n}\frac{1}{k} $, then

$\begin{array}\\ y_{n}-y_{n-1} &=\sum\limits_{k=n+1}^{2n}\frac{1}{k} -\sum\limits_{k=n}^{2n-2}\frac{1}{k}\\ &=\frac1{2n}+\frac1{2n-1}-\frac1{n}\\ &=\frac1{2n}+\frac1{2n-1}-(\frac1{2n}+\frac1{2n})\\ &=\frac1{2n-1}-\frac1{2n}\\ &<\frac1{2n-2}-\frac1{2n}\\ &=\frac12(\frac1{n-1}-\frac1{n})\\ \end{array} $

and $y_{n}-y_{n-1} > 0$.

Summing from $n+1$ to $m$,

$y_m-y_n =\sum\limits_{k=n+1}^m (y_{k}-y_{k-1}) > 0 $ and $y_m-y_n =\sum\limits_{k=n+1}^m (y_{k}-y_{k-1}) <\sum\limits_{k=n+1}^m \frac12(\frac1{k-1}-\frac1{k}) = \frac12(\frac1{n}-\frac1{m}) < \frac1{2n} $.

By the Cauchy criteria, $y_n$ converges, without showing what the limit is.

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