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1 vote
3 answers
117 views

Evaluate $\prod_{n\geq2}\left(\frac{4}{e^2}\left(1+\frac{1}{n}\right)^{2n+1}\frac{n^2-1}{4n^2-1}\right)$

$$\prod_{n\geq2}\left(\frac{4}{e^2}\left(1+\frac{1}{n}\right)^{2n+1}\frac{n^2-1}{4n^2-1}\right)$$ I expand $\frac{n^2-1}{4n^2-1}$ as $\frac{(n-1)(n+1)}{(2n-1)(2n+1)}$. Then, $\left(1+\frac{1}{n}\...
user avatar
3 votes
2 answers
296 views

if $\lim\limits_{n \to \infty} b_n =0 $ then how to prove that $\lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k}=0$

in Problems in Mathematical Analysis I problem 2.3.16 a), if $\lim\limits_{n \to \infty}a_n =a$, then find $\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}$ The proof that ...
pie's user avatar
  • 6,563
2 votes
1 answer
97 views

Find the interval of convergence $\sum_{n\geq1}\left(\ln\frac{1}{2}+1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{(-1)^n}{n}\right)x^n$

Find the interval of convergence A,of the $$\sum_{n\geq1}\left(\ln\frac{1}{2}+1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{(-1)^n}{n}\right)x^n$$ and calculate the sum for each value of $x\in A$ . My work $$...
user avatar
1 vote
1 answer
101 views

Compute ${\sum_{n=1}^{+\infty}(-1)^n\left(\sum_{k=1}^{n}\frac{1}{2k-1}-\frac{ \ln n}{2}-\frac{\gamma}{2}-\ln 2\right)}$

Compute $${\sum_{n=1}^{+\infty}(-1)^n\left(\sum_{k=1}^{n}\frac{1}{2k-1}-\frac{ \ln n}{2}-\frac{\gamma}{2}-\ln 2\right)}.$$ What I have done so far Lemma: $$\displaystyle{\mathop {\lim }\limits_{N \to \...
Mods And Staff Are Not Fair's user avatar
1 vote
0 answers
103 views

Restructuring Jacobi-Anger Expansion

In Jacobi-Anger expansion, $$e^{\iota z \sin(\theta)}$$ can be written as: $$e^{\iota z \sin(\theta)} = \sum_{n=-\infty}^{\infty} J_n(z)e^{\iota n \theta}$$ where $J_n(z)$ is the Bessel function of ...
SiPh's user avatar
  • 31
10 votes
3 answers
614 views

Show that $\sum_{n=1}^{+\infty}\frac{1}{(n\cdot\sinh(n\pi))^2} = \frac{2}{3}\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{(2n-1)^2} - \frac{11\pi^2}{180}$

What I do so far \begin{align*} \text{Show that} \quad &\sum_{n=1}^{+\infty}\frac{1}{(n\cdot\sinh(n\pi))^2} = \frac{2}{3}\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{(2n-1)^2} - \frac{11\pi^2}{180} \\ \...
Mods And Staff Are Not Fair's user avatar
1 vote
3 answers
110 views

Did I solve the problem $\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{{n\choose k}} \to 2$ correctly?

I'm in a calculus class I need to prove that $$\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{{n\choose k}} \to 2$$ Is my below solution valid? We are given the problem $$\lim_{n\to\infty}\sum_{k=0}^{n}\...
Omori's user avatar
  • 133
0 votes
1 answer
37 views

Let $s_n=\sum_l (e^{ix} − 1)^{-1}(e^{inx} − e^{-i(n-1)x})$ where $x=c+l\omega$ . How to show that $s_1+...+s_n=\sum_l \frac{1-\cos nx}{1-\cos x }?$

If i define $s_n=\sum_l (e^{i(c+l\cdot \omega)} − 1)^{-1}(e^{in(c+l\cdot \omega)} − e^{-i(n-1)(c+l \cdot \omega)})$ where $c \in [0,2\pi]$, how can i show that $$s_1+s_2+...+s_n=\sum_l \frac{1-\cos n(...
Rocc_00's user avatar
  • 41
0 votes
1 answer
81 views

Prove that $\int_0^1\lfloor nx\rfloor^2 dx = \frac{1}{n}\sum_{k=1}^{n-1} k^2$

First of all apologies for the typo I made in an earlier question, I decided to delete that post and reformulate it I am asked to prove that $$\int_{(0,1)} \lfloor nx\rfloor^2\,\mathrm{d}x =\frac{1}{n}...
John Doe's user avatar
  • 131
3 votes
2 answers
206 views

why does $\pi$ always show up in $\int_0 ^1 \frac{x^c}{1+x^k} dx$ if $c\neq mk-1$ for all $m \in \mathbb{N}$

when I posted this question I was interested in the sum $ \sum_{n=0} ^{\infty} \frac{(-1)^n}{4n+3}$ but when I thought about the generalised sum $ \sum_{n=0} ^{\infty} \frac{(-1)^n}{kn+c +1}$ for all $...
pie's user avatar
  • 6,563
3 votes
2 answers
161 views

$\sin x \cos x$ defined as a Cauchy Product of their Taylor Series.

The goal of this project is to show that the Cauchy Product of two Taylor Series, $\sin x$ and $\cos x$, is equal to the Maclaurin Series of $\sin x\cos x$ . I am having trouble simplifying the ...
wiley.sg's user avatar
  • 106
1 vote
1 answer
81 views

Help needed in understanding how a certain result was found.

$$\sum_{j=1}^{a-1}\widetilde{\zeta}(j+1)\widetilde{\zeta}(2a-j)=\sum_{j=1}^{a-1}\widetilde{\zeta}(2j)\widetilde{\zeta}(2a+1-2j)$$ I'm trying to understand how the two are equal. At first, I thought ...
Karabo Dibakoane's user avatar
1 vote
1 answer
230 views

Compute $\lim\limits_{n\rightarrow+\infty}(\sum\limits_{i=1}^n(1+\frac{i}{n})^i)^{\frac{1}{n}}$

Here is a question in calculus. Compute the limit of the sequence: $\lim\limits_{n\rightarrow+\infty}(\sum\limits_{i=1}^n(1+\frac{i}{n})^i)^{\frac{1}{n}}$? There are in general three ways to compute ...
Hebe's user avatar
  • 825
11 votes
3 answers
451 views

How to evaluate $ \sum\limits_{k=0} ^{\infty} \frac{(-1)^k}{4k+3}$?

I was trying to solve the integral $\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}}dx$ and I noticed I can do the following: $$\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}}dx=\int_0 ^{\frac{\pi}{4}} \sqrt{\tan{x}} \...
pie's user avatar
  • 6,563
5 votes
2 answers
185 views

Generating Function $\sum_{k=1}^{\infty}\binom{2k}{k}^{-2}x^{k}$

Closed Form For : $$S=\sum_{k=1}^{\infty}\binom{2k}{k}^{-2}x^{k}$$ Using the Series Expansion for $\arcsin^2(x)$ one can arrive at : $$\sum_{k=0}^{\infty}\binom{2k}{k}^{-1}x^{k}=\frac{4}{4-x}-4\arcsin\...
Miracle Invoker's user avatar

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