given the series  $$ \sum_{n=0}^{\infty} \frac{x^{n}}{an+b}=f(x) $$  so for any postive integer $ an+b $ is never 0  how could i get $ f(x) $  from this Taylor series ??
120840
Known that $\sum_{n=0}^{\infty}{x^n}{z^n}=\frac{1}{1-xz}$.  If we have   $\sum_{n=0}^{\infty}\frac{{x^n}{z^n}}{n\beta + \alpha}$ where $\beta, \alpha $ are element of real numbers but not equal $0$. What is a suitable expression for that summation?
94456
What is the radius of convergence of the Taylor series of $\frac{2\tan x}{1+4x^2}$ around $x=0?$  I have to find out $a_n=\frac {f^{(n)}(0)}{n!}$, but the $n-$th  derivative of the given function is not easy to handle. Please help. Thanks,
38634
$$ f: \Re \longrightarrow \Re \ \in C^{\infty} \\ \exists \ L&gt;0: \ \forall x \in \Re, \forall n\in N \\ |f^{(n)} (x)| \le L \\ f(\frac{1}{n})=0 \ \forall n\in N \\ f(x) \equiv 0  $$  Good morning,  Can you give me an help to make this proof?  This is my thought:   $\lim_{n \rightarrow +\infty } \dfrac{1}{n}=0=\theta$ $f(\theta)=0$ $f^{(n)} (\theta) =0\\f(x)=\sum_{k=0}^n \dfrac{f^{(k)}(x-\theta)^k}{k!}$  $f$ is analytic, the rest of Taylor's expansion is 0, because the derivatives are bounded, $f$ is null.   I have thought this reasoning, but I have difficulties to demonstrate part 3. Can you help me?  Is there any mistake?  Can I use an easier way to make the proof?  Thanks.
34337
Compute the taylor series of $\ln(1+x)$  I've first computed derivatives (up to  the 4th) of ln(1+x)  $f^{'}(x)$ = $\frac{1}{1+x}$  $f^{''}(x) = \frac{-1}{(1+x)^2}$ $f^{'''}(x) = \frac{2}{(1+x)^3}$ $f^{''''}(x) = \frac{-6}{(1+x)^4}$ Therefore the series: $\ln(1+x) = f(a) + \frac{1}{1+a}\frac{x-a}{1!} - \frac{1}{(1+a)^2}\frac{(x-a)^2}{2!} +  \frac{2}{(1+a)^3}\frac{(x-a)^3}{3!} -  \frac{6}{(1+a)^4}\frac{(x-a)^4}{4!} + ...$  But this doesn't seem to be correct. Can anyone please explain why this doesn't work?  The supposed correct answer is:  As $\ln(1+x) = \int (\frac{1}{1+x})dx$ $\ln(1+x) = \Sigma_{k=0}^{\infty} \int (-x)^k dx$
38229
Given the series $S=\sum_{n=1}^{\infty} \frac{(2n)!}{4^n(n!)^2}$ I'm trying to prove its divergent but with no luck. It doesn't tend to $\infty$ as $n$ grows large and ratio test is inconclusive. I have been trying to do a comparison test but all series less than $S$ that I was able to come up with were convergent. So I'm starting to suspect that the series might actually be convergent, any hints?  Thanks!
76955
I have figured out a way to show that     $$e=\sum_{i=0}^\infty {1\over i!}$$   I am wanting to formally show that     $$e^x = \sum_{i=0}^\infty {x^i\over i!}$$   I have been looking at power series/Taylor series for a long period of time (absolute convergence) and have seen multiple proofs that I look past because something seems illegitimate with radius of convergence. If someone explains the general proof behind Taylor series/power series absolutely converging so there is no gray area, that would work as well. I will probably ask some questions in the comment area if this is the case. There must be something I am missing.
5245
I have to find the sum of the following series:  $$\sum\limits_{n=1}^\infty \frac{n^3}{n!}$$  I know how to prove the convergence of this series, but how do i find the sum. I can't use the properties of derivation or integration of the series, i have to do it using the definition that a series converges iff it's sequence of partial sum converges.
36879
Does the following series converge:  $$\sum_{n=3}^\infty \frac {1}{(\log n)^{\log(\log(n)}}$$  I've tried all test I know... Any ideas ?
57644
Finding sum of series $\displaystyle \sum^{\infty}_{k=0}\frac{(k+1)(k+2)}{2^k}$   what i try  Let $\displaystyle x=\frac{1}{2},$ Then series sum is $\displaystyle \sum^{\infty}_{k=0}(k+1)(k+2)x^k$  $\displaystyle \sum^{\infty}_{k=0}x^k=\frac{1}{1-x}\Rightarrow \sum^{\infty}_{k=0}x^{k+1}=\frac{x}{1-x}=\frac{1-(1-x)}{1-x}$  differentiating both side  $\displaystyle \sum^{\infty}_{k=0}(k+1)x^{k}=+\frac{1}{(1-x)^2}$  How do i solve it help me please
56039
I think that the series $$\sum_{n=1}^{\infty} \dfrac{\sin(n)}{n}$$ converges conditionally, but I´m not able to prove it. Any suggestions ?
34077
Given the Taylor series $\sum a_k (x - x_0)^k$ of an analytic function, it is possible to determine whether the function is periodic more-or-less directly from the coefficients $a_0, a_1, \ldots$ of the series (equivalently, the derivatives $f'(x_0), f''(x_0), \ldots$ of $f$ at $x_0$)?  (For simplicity, I'm happy to restrict if necessary to the case that the series converges on all of $\mathbb{R}$. Note too that it's no real loss of generality to take $x_0 = 0$.)  (This was motivated by .)
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.
75893
This is the sum   $$\sum_{n=1}^{\infty}\frac{1}{{5^n}^!}$$  My first attempt was to assume that the series does converge to a rational number $a/b$. But the $n!$ bothered me and I failed in my proof.  How would you try to prove this series?
95818
$$\sum_{n=0}^\infty n!x^{n^2}$$   Can anyone tell me what will be the radius of convergence for this power series. According to me it is $0$.   If you can please show me how to go about it if I am wrong.
13007
What is the sum of $$\sum_{n=0}^{\infty}\frac{1}{2^n(n+1)}$$    I've spent an insane amount of time on this problem. I checked on Wolfram and it gives $ln(4)$, which I assume you get from $2\ln(2)$.   I know that the integral of $\frac{1}{n+1}$ is $\ln(n+1)$. What am I missing?  Thank you!
38303
As the title says, given a series $b_n &gt; 0$, where $\sum_{n=1}^\infty b_n$ is divergent:  Show that the series $$\sum_{n=1}^\infty \frac{b_n}{1+b_n}$$ is also divergent.  So I've defined the series as $\sum_{n=1}^\infty u_n v_n$ where $u_n = b_n$ and $v_n = \frac {1}{1+b_n}$.   I'm sure there was some sort of convergence test which stipulated that for a product of two series, if one diverges the entire thing does, however I might be mistaken. If so, how do I properly prove this?
99041
Find $f(x)$, the unknown function satisfying   $$f(x) = \sum\limits_{n=0}^{+\infty}\frac{x^{4n}}{(4n)!}$$   I'm looking for a direct solution which is different from mine, if possible.