If $x\notin \mathbb{R}$ then $$\sum_{k=0}^{n-1}\sin(2\pi kx)=\frac{\sin((n-1)\pi x)\sin(n\pi x)}{\sin(\pi x)}$$ This supposedly is a direct results from Dirichlet kernel. I was also told that this can simply found by using additions of sine functions, but I'm at a loss. Any help is appreciated.
1806
How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? For example here is the sum of $\cos$ series:  $$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$  There is a slight difference in case of $\sin$, which is: $$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2  a + (n-1)\cdot d}{2}\biggr)$$   How do we prove the above two identities?
117534
Consider the following series $$S_N = \sum_{k=1}^N \frac{\sin[k\pi/(N+1)]}{k}.$$ What is the asymptotic behavior of this series for $N \to \infty$?  If we replace the sine function with the factor $1$, we get the standard harmonic series which diverges $\sim \ln N$. However, if we replace the sine function with $(-1)^k$, we get a convergent series with the result $\ln 2$. Therefore, I think that $S_N$, defined above, should diverge, but more slowly than the logarithm. Am I correct?
98278
$$\sum_{r=0}^{k-1}4^r$$ Hi, I was wondering whether anyone could explain how to work this out. I know the end result is $\frac{4^k-1}{3}$, but I don't know why or how to get there.  Thank you :D
17273
I was asked to evaluate the following sum: $$\sum_{k=1}^{\infty}\frac{1}{k}\int_{\pi k}^{\infty}\frac{\sin(x)}{x}dx$$  I'm trying to use $$\int_{0}^{\infty} \frac{\sin(x)}{x}dx=\frac{\pi}{2}$$ However, it doesn't seem to work. Any help is greatly appreciated.
77797
Compute $$\displaystyle \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^n (\sin{x})^kdx.$$   I tried writing $\displaystyle \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^n (\sin{x})^kdx = \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \frac{1-(\sin{x})^{n+1}}{1-\sin{x}}dx$, but I don't know how to continue form here.  I know that the limit diverges to $\infty$, but I don't know how to prove it.
64120
I recently came across this problem:  Prove that $\int_{0}^{\frac{\pi}{2}} \sin(\sin y) dy \leq 1$.  However, from my understanding, the expression $\sin(\sin y)$ does not have an indefinite integral (). So, how will we go about solving the above problem using just the Riemann sums?
25401
For any $x\in\mathbb{R}$, the series $$ \sum_{n\geq 1}\tfrac{1}{n}\,\sin\left(\tfrac{x}{n}\right) $$ is trivially absolutely convergent. It defines a function $f(x)$ and I would like to show that $f(x)$ is unbounded over $\mathbb{R}$. Here there are my thoughts/attempts:   $$(\mathcal{L} f)(s) = \sum_{n\geq 1}\frac{1}{1+n^2 s^2} = \frac{-s+\pi\coth\frac{\pi}{s}}{2s}=\sum_{m\geq 1}\frac{(-1)^{m+1}\,\zeta(2m)}{s^{2m}}$$ is a function with no secrets. It behaves like $\frac{\pi}{2s}$ in a right neighbourhood of the origin, like $\frac{\pi^2}{6s^2}$ in a left neighbourhood of $+\infty$. The origin is an essential singularity and there are simple poles at each $s$ of the form $\pm\frac{i}{m}$ with $m\in\mathbb{N}^+$. These facts do not seem to rule out the possibility that $f$ is bounded; For any $N\in\mathbb{N}^+$ there clearly is some $x\ll e^N$ such that $\sin(x),\sin\left(\frac{x}{2}\right),\ldots,\sin\left(\frac{x}{N}\right)$ are all positive and large enough, making a partial sum of $ \sum_{n\geq 1}\tfrac{1}{n}\,\sin\left(\tfrac{x}{n}\right) $ pretty close to $C\log N$. On the other hand I do not see an effective way for controlling  $ \sum_{n&gt;N}\tfrac{1}{n}\,\sin\left(\tfrac{x}{n}\right) $ - maybe by summation by parts, by exploiting the bounded-ness of the sine integral function? Some probabilistic argument might be effective. For any $n\geq 3$ we may define $E_n$ as the set of $x\in\mathbb{R}^+$ such that $\sin\left(\frac{x}{n}\right)\geq \frac{1}{\log n}$. The density of any $E_n$ in $\mathbb{R}^+$ is close to $\frac{1}{2}$, so by a Borel-Cantellish argument it looks reasonable that the set of points such that $|f(x)|\geq \frac{\log x}{100}$ is unbounded, but how to make it really rigorous? To compute $\lim_{x\to x_0}f(x)$ through convolutions with approximate identities seems doable but not really appealing.
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.
107807
I was wondering if there is a closed-form Laplace inverse of the sine function. I have tried the following:  $$   \sin(as)=\sum_{n=0}^{\infty}\frac{(-1)^{n}(as)^{2n+1}}{(2n+1)!} $$ an $n$-th power of $s$ contributes with an $n$-th derivative of the Dirac delta. So one expects a series expansion in terms of the Delta function and its derivatives. But that is utterly ugly! Hence the question.
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?
4704
Other than the general inductive method,how could we show that $$\sum_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$$  Apart from induction, I tried with  to check the validity, but I can't think of an easy (manual) alternative.  Please suggest an intuitive/easy method.
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?
12612
For what values of $a$ does $$\sum_{n=1}^\infty \left( 1+\frac12 +  \dotsb + \frac1n \right) \frac{\sin (na)}{n}$$ converge?   To my thinking, $$f(n)=\frac{\left( 1+\frac12 +  \dotsb + \frac1n \right)}{n}$$ will behave like $\frac{\log n}{n}$, which is not convergent, but certainly has terms going to zero (eventually monotonically). By Dirichlet's test, $$\sum_{n=1}^\infty f(n) \sin(an)$$ will converge provided that the partial sums of $\sum_{n=1}^\infty \sin(na)$ are bounded, which they always are. So it should converge for all $a\in \mathbb{R}$.  Would appreciate anyone pointing out mistakes in this reasoning. Thanks
6333
I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:  $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$  I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
100910
I was stuck while finding the given summation.  $$\sum_{k=1}^rk^2\binom {n-k}{r-k}$$  Since $n$ and $r$ are both constants, so I have first converted the above summation into this:  $$\sum_{k=1}^rk^2\binom {n-k}{n-r}$$  I have no idea how to proceed next. Any help will be appreciated.
98662
At  I've read that being $\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$ by using  we get $\sin'(x)=\sum_{n=0}^{\infty}(2n+1) \frac{(-1)^n}{(2n+1)!}x^{2n}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=\cos(x)$ but the page  says that if $f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n$ then inside the radius of convergence of the power series we have $f'(x)=\sum_{n=1}^{\infty}nc_n(x-a)^{n-1}$ so shouldn't it be $\sin'(x)=\sum_{n=1}^{\infty}(2n+1)\frac{(-1)^n}{(2n+1)!}x^{2n+1-1}=\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}$?  Now I understand that this last result has a missing term due to the index of the series being $1$ and not $0$ but being the $1$ mandatory because of the above theorem and since trying to shift index doesn't help (we get $\sin'(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(2n+2)!}x^{2n+2}$) how can the result that ProofWiki shows be rigorously justified?   (I mean, I think we can't just say "oh, there's a missing term!" after having applied the theorem and magically shift the index of the series to obtain the desired result)
89519
So I have been trying for a few days to figure out the sum of   $$ S = \sum_{k=1}^\infty \frac{1}{k^2 - a^2} $$ where $a \in (0,1)$. So far from my nummerical analysis and CAS that this sum equals  $$ S = \frac{1}{2a} \left[ \frac{1}{a} \, - \, \pi \cot(a\pi) \right] $$  But I have not been able to prove this yet. Anyone know how? My guess is that the  sum of this series is related to fourier-series but nothing particalr comes to mind.   For the easy values, I have been able to use telescopic series, and a bit of algebraic magic, but for the general case I am stumpled. Anyone have any ideas or hints? Cheers.