I'm taking a single-variable calculus course and asked to calculate $$\sum_{k=0}^\infty (-1)^k \frac{\pi^{2k}}{(2k)!}$$ So the $k=0$ term looks like $$(-1)^0 \frac{\pi^{2(0)}}{(2(0)!)}=\frac{1}{0!}$$ but $\frac{1}{0!}$ is not a number, so it appears I've made a mistake in my calculation.
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How can I prove  from first principles that $0!$ is equal to $1$?
97412
I'm looking for a way to prove $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$$ I know that $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{1}{4^{2m+1}}\left(\zeta\left(2m+1,\frac14\right)-\zeta\left(2m+1,\frac34\right)\right)$$ so maybe I could simplify the above more?
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?
88825
$\sum_{n=1}^\infty n*(n+1)*x^n$  Hello everyone,  I need help in solving the question above.  I started with $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$, if differentiated it once so it became $\frac{1}{(1-x)^2} =\sum_{n=1}^\infty nx^{n-1} $ but from here I don't know how to continue. Thanks.
52326
Show that: $$\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=e-1$$  First I will re-define the sum: $$\sum^\infty_{n=1} \frac{n^2}{(n+1)!} = \sum^\infty_{n=1} \frac{n^2-1+1}{(n+1)!} - \sum^\infty_{n=1}\frac{n-1}{n!} + \sum^\infty_{n=1} \frac{1}{(nm)!}$$  Bow I will define e: $$e^2 = 1+ \frac{2}{1!} + \frac{x^2}{2!} + ... + \infty$$ $$e' = 1 + \frac{1}{1!} + \frac{1}{2!} + ... + \infty$$ $$(e'-2) = \sum^\infty_{n=1} \frac{1}{(n+1)!}$$  Now I need help.
99071
I typed into my Mathematica:$\sum _{n=1}^{\infty } \frac{1}{n^2+1}$ , and the result was: $$\frac{-1+\pi \coth (\pi)}{2}$$ I know how to estimate the aforementioned sum , but I have no idea how to get its closed form.
114767
First I have checked the search option but found nothing relevant to my problem and also level of math. I just started learning the language of mathematics, on my own and I have trouble understanding why  $  \frac{1}{\infty } \approx 0 $ and  why $\frac{1}{0} = {\infty}$?  I want to know how exactly do I get to these answers. It might sound strange but it's hard for me as a beginner to picture things like this on my own.
39958
I have this problem :  $$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1}$$  My solution  $$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1} = -\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k+1} -\frac{1}{k-1} = -\frac{1}{2}[(\frac{1}{3}-1)+(\frac{1}{4}-\frac{1}{2})+(\frac{1}{5}-\frac{1}{3})+(\frac{1}{6}-\frac{1}{4})+...]$$  I think that the sum should be $\frac{1}{2}$ since the limit of :  $$-\frac{1}{2}[-1+\frac{1}{k}+...+\frac{1}{n}] = \frac{1}{2}$$  But that wrong. Any ideas?
95733
Calculate $$\sum \limits_{k=0}^{\infty}\frac{1}{{2k \choose k}}$$  I use software to complete the  series is  $\frac{2}{27} \left(18+\sqrt{3} \pi \right)$  I have no idea about it.  :|
81224
What is the value of this : $$\int_{0}^{1}x^{{(\frac{1}{\sum_{a=0}^{\infty}\sum_{b=0}^{\infty}\sum_{c=0}^{\infty}\frac{1}{(a+b+c)!}})}x^{(\frac{1}{\sum_{a=0}^{\infty}\sum_{b=0}^{\infty}\sum_{c=0}^{\infty}\frac{1}{(a+b+c)!}})}}dx$$ My work: we have (1)..$$\sum_{a=0}^{\infty}\sum_{b=0}^{\infty}\sum_{c=0}^{\infty}\frac{1}{(a+b+c)!}=\frac{7e}{2}$$ and : $$\int_{0}^{1}x^{\alpha x^{\beta}}dx=\sum_{n=0}^{\infty}\frac{\alpha^n(-1)^n}{{(\beta n+1)}^{n+1}}$$ So $$\int_{0}^{1}x^{{(\frac{1}{\sum_{a=0}^{\infty}\sum_{b=0}^{\infty}\sum_{c=0}^{\infty}\frac{1}{(a+b+c)!}})}x^{(\frac{1}{\sum_{a=0}^{\infty}\sum_{b=0}^{\infty}\sum_{c=0}^{\infty}\frac{1}{(a+b+c)!}})}}dx=\int_{0}^{1}x^{(\frac{2}{7e}) x^{(\frac{2}{7e})}}dx=\sum_{n=0}^{\infty}\frac{(\frac{2}{7e})^n(-1)^n}{{((\frac{2}{7e}) n+1)}^{n+1}}$$ But$$\sum_{n=0}^{\infty}\frac{(\frac{2}{7e})^n(-1)^n}{{((\frac{2}{7e}) n+1)}^{n+1}}=?$$ You can help calculate the series or find an integral value in another way
9386
Can you please explain why $$ \sum_{k=1}^{\infty} \dfrac{k}{2^k} = \dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots = 2 $$  I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
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I guess the proof of the identity $$ \sum_{n = 0}^{\infty} \frac{1}{n!} \equiv \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x $$ explains the connection between such different calculations. How is it done?
27856
A previous problem had us solving $\sum_{n=0}^{\infty} \frac{1}{4^n}$ which I calculated to be $\frac{4}{3}$ using a bit of mathematical manipulation.   Wonderful. Thank you for all the prompt responses.  Could anyone suggest an alternate technique that does not involve differentiation?
66506
How can I calculate $\sum\limits_{n=1}^{\infty}\frac{1}{F_n}$, where $F_0=0$, $F_1=1$ and $F_n=F_{n-1}+F_{n-2}$?  Empirically, the result is around $3.35988566$.  Is there a "more mathematical way" to express this?
67934
$$\frac{\sum_{i=0}^nx_iy_i-\bar y\sum_{i=0}^nx_i}{\sum_{i=0}^nx_i^2-\bar x\sum_{i=0}^nx_i}= \frac{\sum_{i=0}^n(x_i-\bar x)(y_i-\bar y)}{\sum_{i=0}^n(x_i-\bar x)^2}$$  Maybe it refers some calculation rules about summation. I cannot solve that. Thanks!
68613
$\sum_{n=0}^{\infty}z^{n!}$  My textbook says its answer is 1. But I think the sequence is not power series because it cannot be expressed as the form of $CnZ^n$.   Am I right? Or where do I have a mistake?
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.