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How can I find the limit of

$$ u_n =\sin\left(\frac{1}{n+1}\right)+\cdots+\sin\left(\frac{1}{2n}\right)$$

when $n\rightarrow\infty$?

We have:

$$ \sum_{n=1}^\infty u_{n+1}-u_n =u_\infty -\sin\left(\frac{1}{2}\right)$$

So how can I find $$ \sum_{n=1}^\infty u_{n+1}-u_n =\sum_{n=1}^\infty \sin\left(\frac{1}{2n+2}\right)+\sin\left(\frac{1}{2n+1}\right)-\sin\left(\frac{1}{n+1}\right)\ ?$$

2 Answers 2

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We can make use of the fact that $$\sin x = \sum\limits_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ and note that when $x\leq 1$, the absolute value of each term is more than the sum of the absolute values of the later terms. Thus $x-\frac{x^3}{6}<\sin x for $x\leq 1$, so we have $$\sum\limits_{k=n+1}^{2n}\frac{1}{k}-n\frac{1}{6n^3}<\sum\limits_{k=n+1}^{2n}\frac{1}{k}-\frac{1}{6k^3} and $$\begin{eqnarray}\lim\limits_{n\to\infty}\sum\limits_{k=n+1}^{2n}\frac{1}{k}&=&\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^{2n}\frac{1}{k}-\ln(2n)\right)-\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^{n}\frac{1}{k}-\ln(n)\right)+\lim\limits_{n\to\infty}(\ln(2n)-\ln(n))\\ &=&\lim\limits_{n\to\infty}(\ln(2n)-\ln(n))\\ &=&\lim\limits_{n\to\infty}\ln 2\\ &=&\ln 2\end{eqnarray}$$ while clearly $\lim\limits_{n\to\infty}n\frac{1}{6n^3}=0$, so by squeeze theorem $\lim\limits_{n\to\infty} u_n=\ln 2$.

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    Same ideas but different details: We can get the required inequality more simply and without condition like [this](http://math.stackexchange.com/a/111914/14657). And $\displaystyle \lim_{n\to \infty} \frac{1}{n} \sum_{k=n+1}^{2n} \frac{1}{(k/n)} = \int^2_1 \frac{1}{t} dt = \log 2.$2012-02-26
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    @RagibZaman Ah, very nice use of Riemann sums.2012-02-26
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Using $$ x - \frac{x^3}{6} \lt \sin x \lt x$$

for positive $x$ close to $0$ and the fact that $\displaystyle \sum_{k=1}^{n} \frac{1}{k^3}$ is convergent, we see that

your limit is the same as the limit of $\displaystyle s_n = \sum_{k=1}^{n} \frac{1}{n+k}$

One method to find this limit is to use the Harmonic series estimate as in Alex's answer.

Another method is to interpret it as a Riemann sum of the integral $\int_{0}^{1} \frac{1}{1+x} \text{ dx}$ as in Ragib's comment to Alex's answer.

I will mention a third method, which uses the Mean Value theorem. (though all three are quite similar).

Since $\frac{1}{x}$ is decreasing for positive $x$, we see using that mean value theorem that

$$\frac{1}{x+1} \lt \log (1+x) - \log x \lt \frac{1}{x}$$

Setting $x = n+1, n+2, \dots, 2n$ and adding, and then setting $x=2n-1, 2n-2, \dots, n$ and adding we get

$$\log \left(\frac{2n+1}{n}\right) \lt s_n \lt \log \left(\frac{2n-1}{n-1}\right)$$

and thus

$$\lim_{n \to \infty} s_n = \log 2$$