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I stuck with these 2 limits, can you help me please?

$1.\displaystyle\quad \lim_{n \to \infty }\frac{\sin1+2\sin\frac{1}{2}+3\sin\frac{1}{3}+\cdots+n\sin\frac{1}{n}}{n}$

$2.\displaystyle\quad \lim_{n \to \infty }\frac{n}{\frac{1}{\sin1}+\frac{1/2}{\sin1/2}+\frac{1/3}{\sin1/3}+\cdots+\frac{1/n}{\sin1/n}} $

Thanks in advance.

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try to prove: if $\lim_{n\to\infty}a_n=1$ then $\lim(a_1+\dots+a_n)/n=1$ – user8268 Nov 6 '12 at 8:53
@user8268 This is exactly the point where i stuck, how can I to prove that $\sum_{i=1}^{n}n\sin\frac{1}{n}=1$? – Tina Nov 6 '12 at 9:35
@Tina : It's not true that $\displaystyle\sum_{i=1}^n n\sin\frac1n=1$. But it is true that $\displaystyle\lim_{n\to\infty}n\sin\frac1n=1$. That's what user8268 was suggesting you use. – Michael Hardy Nov 6 '12 at 10:29
The first one:… – Martin Sleziak Sep 3 '14 at 14:45
up vote 3 down vote accepted

We will use unnecessarily explicit inequalities to prove the result.

In the first limit, the general term on top can be rewritten as $\dfrac{\sin(1/k)}{1/k}$. This reminds us of the $\frac{\sin x}{x}$ whose limit as $x\to 0$ we needed in beginning calculus.

Note that for $0\lt x\le 1$, the power series $$x-\frac{x^3}{3!}+\frac{x^5}{5!} -\frac{x^7}{7!}+\cdots$$ for $\sin x$ is an alternating series. It follows that for $0\lt x\le 1$, $$x-\frac{x^3}{6}\lt \sin x\lt x.$$ and therefore $$1-\frac{x^2}{6}\lt \frac{\sin x}{x}\lt 1.$$ Put $x=1/k$. We get $$1-\frac{1}{6k^2}\lt \frac{\sin(1/k)}{1/k} \lt 1.\tag{$1$}$$ Add up, $k=1$ to $k=n$, and divide by $n$ Recall that $$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots =\frac{\pi^2}{6}.$$ We find that $$1-\frac{\pi^2}{36n}\lt \frac{\sin1+2\sin\frac{1}{2}+3\sin\frac{1}{3}+\cdots+n\sin\frac{1}{n}}{n}\lt 1.$$ From this, it follows immediately that our limit is $1$.

A very similar argument works for the second limit that was asked about. It is convenient to consider instead the reciprocal, and calculate $$\lim_{n \to \infty }\frac{\frac{1}{\sin1}+\frac{1/2}{\sin1/2}+\frac{1/3}{\sin1/3}+\cdots+\frac{1/n}{\sin1/n}}{n}.$$ We can then use the inequality $$1\lt \frac{1/k}{\sin(1/k)} \lt \frac{1}{1-\frac{1}{6k^2}},$$ which is simple to obtain from the Inequalities $(1)$. Having the $1-\frac{1}{6k^2}$ in the denominator is inconvenient, so we can for example use the inequality $\dfrac{1}{1-\frac{1}{6k^2}}\lt 1+\dfrac{1}{k^2}$ to push through almost the same proof as the first one.

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As $n \to \infty , \sin(1/n)\to 1/n , n\sin(1/n)\to 1$. Now, by Cesaro mean

$\lim\limits_{n \to \infty} \sum_{1}^{n}n\sin(1/n)\to n$.

Distributing the it over numerator and denominator

$\lim\limits_{n \to \infty} \frac{\sum_{1}^{n}n\sin(1/n)}{n}= \frac{\lim\limits_{n \to \infty }n\sin(1/n)}{n}=1$

So, the answer to the first part is 1 . Same argument holds for the second part too.

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$\sum_{1}^{n}n\sin(1/n)$ is not a very good notation, since you use $n$ both in range and as a variable. I think that $\sum_{k=1}^{n}k\sin(1/k)$ would be better. – Martin Sleziak Nov 6 '12 at 10:43
Thanks for the edit Martin. This was my first answer at math stackexchange. Will keep it in mind. – dexter04 Nov 7 '12 at 9:24

the first one , you can use the stolz theorem directly. or use the result: if $\lim\limits_{n\to \infty}a_n=a$, then $\lim\limits_{n\to \infty}\frac{a_1+a_2+.....a_n}{n}=a$, you can use the $\epsilon-N$ to illustrate it...

the second is the same

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This is exactly the point where i stuck, how can I to prove that $\sum_{i=1}^{n}n\sin\frac{1}{n}=1$?Thanks. – Tina Nov 6 '12 at 9:37
your above gauss is wrong because $\lim\limits_(n\sin\frac{1}{n})\to 1$, therefore, the series cannot be convengent.... – Tao Nov 6 '12 at 10:44

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