We need to calculate this:  $$\lim_{n \to \infty} { \sum_{k=1}^{n} { (\frac{nk-1}{n^3})\sin(\frac{k}{n}) }       }$$  So I know this is Riemann sum. This is what I started doing:     $$\sum_{k=1}^{n} { (\frac{nk-1}{n^3})\sin(\frac{k}{n})}  = \sum_{k=1}^{n} { (\frac{k}{n^2 } - \frac{1}{n^3})\sin(\frac{k}{n})}$$   But then I noticed that we can not get $\frac{1}{n}$ outside. How do we exactly know what is the partition?
122476
Calculate the limit $$\lim_{n\to \infty} {\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}}$$   How exactly do we calculate this limit of the Riemann sum? I am never able to find what is the partition. I know that our $f(x)$ is $\sin(x)$.
33042
Inspired by  (and far more straightforward, I am guessing), Mathematica tells us that $$\sum_{k=1}^{\infty}\dfrac{{\sin(k)}}{k}$$ converges to $\dfrac{\pi-1}{2}$.   Presumably, this can be derived from the similarity of the Leibniz expansion of $\pi$ $$4\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{2k-1}$$to the expansion of $\sin(x)$ as $$\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)!}x^{2k+1},$$ but I can't see how...  Could someone please explain how $\dfrac{\pi-1}{2}$ is arrived at?
84235
I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:  \begin{align*} \lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)&amp;=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{n}{n^2+k^2}\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{n^2}{n^2+k^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{1}{1+(k/n)^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}f(\frac{k}{n})(\frac{k-(k-1)}{n})\\ &amp;=\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4} \end{align*}  where $f(x)=\frac{1}{1+x^2}$. Is this correct, are there any steps where I am not clear?
121100
While Solving some integral problem, I encountered the following infinite series:  $$\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$$  I have tried many methods including partial fractions... I seek help! Please provide hints if you don't have the complete answer.
75102
I have to calculate the sum of this series $$ \sum_{k=1}^n\sin{k\theta} $$ I tried solving it like this $$ \sum_{k=1}^n\sin{k\theta} = \operatorname{Im}\sum_{k=1}^ne^{(i\theta)k} $$ I recognized it as a geometric sum, but to get the sum I first had to check $$ \left|e^{i\theta}\right| = 1 $$ which means that the geometric sum is divergent. Apparently that is not the right answer, but I can't find the spot where I made the mistake. Could you help me finding it?
122827
Question: Find the sum of the series: $$\lim_{n \to \infty}\frac{\sin1}{1}+\frac{\sin2}{2}+\frac{\sin3}{3}+...+\frac{\sin n}{n}$$  I have no clue how to find this. Obviously I can see the sum will be convergent as the denominator gets increasingly bigger while the numerator is bound between $1$ and $-1$.
113177
assume  $$x_n=\frac{n+1}{2^{n+1}}\sum_{k=1}^n\frac{2^k}{k} ,n=1,2,.....$$  how compute $\lim_{n\to +\infty}x_n$?  Thanks for any hint
23507
I just came across the problem of what happens in the boundary cases of .  More concisely, for what $\theta$ will the following sum converge? $$\sum_{n=1}^{\infty} \frac{e^{ni\theta}}{n}$$ It is well known that this sum will diverge for $\theta=0$ and will converge for $\theta=\pi$. In fact, for any odd multiple of an even root of unity, since we have symmetry, for each line through the origin of the complex plane, we can use the alternating series theorem to show that they converge.  Another idea I had was to split  the sum into imaginary and real parts, and if we can prove each of these converges, then we are all set. So now we must prove that both of the following converge $$\sum_{n=1}^{\infty}\frac{\cos(n\theta)}{n}\space, \space \sum_{n=1}^{\infty}\frac{\sin(n\theta)}{n}$$ Now these can be shown to converge easily by Dirichlet's test for any rational multiple of pi, but what about irrational multiples of pi? I know it has been shown to converge with $\theta=1$, but I don't see how to generalize this.
17841
I tried and got this   $$e=\sum_{k=0}^\infty\frac{1}{k!}=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}$$ $$n!\sum_{k=0}^n\frac{1}{k!}=\frac{n!}{0!}+\frac{n!}{1!}+\cdots+\frac{n!}{n!}=m$$ where $m$ is an integer. $$\lim_{n\to\infty}n\sin(2\pi en!)=\lim_{n\to\infty}n\sin\left(2\pi n!\sum_{k=0}^n\frac{1}{k!}\right)=\lim_{n\to\infty}n\sin(2\pi m)=\lim_{n\to\infty}n\cdot0=0$$  Is it correct?
84834
I need to compute $$\sum_{n=1}^{+\infty} \frac{1}{n^3\sin(n\pi\sqrt{2})}.$$ This an exercice of "Amar and Matheron, complex analysis". I proved the convergence and now to compute the sum, I follow the hint of the book which is : Consider integrals of the form $$\int_{\gamma}\frac{dz}{z^3[\sin(\pi z)\sin(\sqrt{2}-1)\pi z]}$$ for a well-chosen $\gamma.$ I know this a residue theorem application but it seems a bit hard to have the good idea. I also tried with a summation factor. Any help will be greatly appreciated.
56816
$$ \sum_{n=1}^{\infty} \frac{(\sin n+2)^n}{n3^n}$$  Does it converge or diverge?   Can we have a rigorous proof that is not probabilistic?   For reference, this question is supposedly a mix of real analysis and calculus.
113007
Find the limit $$\lim_{n\to \infty}\sum_{k=1}^{n}\left(\frac{k}{n^2}\right )^{\frac{k}{n^2}+1}$$ Have no idea.
113978
I am having trouble seeing how $\sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)}$ equals $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. I can see $\sum_{n=1}^{\infty}\frac{1}{2n+1}+\frac{1}{2n+3}$ but not $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. All help is appreciated, thanks.
101295
I need to study this sum: $\sum_{n=1}^{\infty} \frac{n!\, i^n}{n^n}$. Taking: $$\lim_{n\rightarrow \infty} \; \left|\left(\frac{n!\: i^n}{n^n}\right)^{\frac{1}{2}}\right|$$ $$\rightarrow \lim_{n\rightarrow \infty}  \left|\left(\frac{n!}{n^n}\right)^{\frac{1}{2}}\right|$$  Using Stirling approximation considering the limit:  $$\ln n! \approx n \ln\:n-n$$ then $n! \approx (\frac{n}{e})^n$ (is this correct?):  $$\rightarrow \lim_{n\rightarrow \infty} \left|\frac{n}{n\, e}\right| = \frac{1}{e}$$ This doesn't make too much sense because I know the series diverges.  I think I'm missing a $\sqrt{2\pi n}$ in Stirling approximation but I don't understand why that pops out taking the $e^{(\;)}$ from the first expression.
19745
Prove that : $\displaystyle \lim_{n\to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=\ln 2$    the only thing I could think of is that it can be written like this :  $$ \lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k+n} =\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\frac{k}{n}+1}=\int_0^1 \frac{1}{x+1} \ \mathrm{d}x=\ln 2$$  is my answer right ? and are there any other method ?(I'm sure there are)
51080
I came across the following sum: $$\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}\frac{4^k}{{2k \choose k}}$$    I thought that this can be evaluated using the expansion of $\dfrac{\sin^{-1}x}{\sqrt{1-x^2}}$ but I couldn't make any use of it. Then I tried to use the following: $$\frac{1}{(2k+1)}\frac{1}{{2k \choose k}}=\frac{\Gamma(k+1)\Gamma(k+1)}{\Gamma(2k+2)}=\int_0^1 x^k(1-x)^k\,dx$$ but this didn't help either and now I am stuck.   Any help is greatly appreciated. Thanks!    EDIT:  The sum originated from the following definite integral:  $$\int_0^{\pi/2}\tan^{-1}(\sin x)\,dx$$
84611
I want to evaluate this sum:  $$ \sum^\infty_{k=1} \frac{1}{k(k+p)}$$  My first thought was to consider the partial sums, and after taking out a factor of $1/p$ from $\sum^n_{k=1} 1/k(k+p)$, I get:  $$\frac{1}{p}\left( 1 - \frac{1}{1+p} + \frac{1}{2} - \frac{1}{2+p} + \ldots + \frac{1}{n} - \frac{1}{n+p}\right)$$  As $n \to \infty$, it seems like the series in the parentheses converges to the $p^{th}$ harmonic number, so the series overall would converge to $\frac{H_p}{p}$.  But how could I prove that? I only noticed that this seemed to be that case after trying out a few numbers. Is it even actually true?