I found this interesting function defined by the following infinite series:  $$g(x)=\sum_{n=0}^\infty(-1)^nx^{2^n}$$  And I noticed that it followed the functional equation $g(x)=1-g(x^2)$.  I was interested, particularly, in the following limit:  $$L=\lim_{x\to1^-}g(x)$$  And assuming it exists, the functional equation yields  $$L=1-L\implies L=\frac12$$  Which, at first look, the limit appears to exist.    However, upon closer inspection, convergence gets a little sketchy...    Can we determine if the limit converges?
123242
I want to calculate the limit of this sum :  $$\lim\limits_{x \to 1} {\left(x - x^2 + x^4 - x^8 + {x^{16}}-\dotsb\right)}$$  My efforts to solve the problem are described in the .
19474
Let $(x_n) _{n\ge 1}$ be a sequence of real numbers such that $\lim\limits_{n\to \infty} \frac{x_1^2+x_2^2+...+x_n^2}{n}=0$. Prove that $\lim\limits_{n\to \infty} \frac{x_1+x_2+...+x_n}{n}=0$. I used the definition of the limit to conclude that $\exists N\in \mathbb{N} $ such that $|\frac{x_1^2+x_2^2+...+x_n^2}{n}|&lt;\frac{1}{n^2}$, $\forall n\ge N$. Hence, we get that $|x_1^2+x_2^2+...+x_n^2|&lt;\frac{1}{n}$. Now here comes the part where I am not really sure. I think that this implies that $\sum_{n=1}^{\infty}x_n^2=0$, and as a result $x_n\to 0$,which solves the problem because if we use the Stolz-Cesaro lemma we get that $\lim\limits_{n\to \infty} \frac{x_1+x_2+...+x_n}{n}=\lim\limits_{n\to \infty} x_n=0$.
84143
Consider the series $ \sum_{n=1}^\infty x^2+ n/n^2$  . Pick out the true statements: (a) The series converges for all real values of $x$. (b) The series converges uniformly on $\mathbb{R}$. (c) The series does not converge absolutely for any real value of $x$.        stuck on this problem totally. please help me to solve this problem.thanks .
97976
the series $a_n$ is any arbitrary converging series.  My thought process was that the $1/n$ will definitely go to zero as n approaches infinity; however, the series $k*a_k$ seems to approach infinity at the same rate. This confuses me because if both the numerator and denominator approach infinity, I think that the overall equation will equal 1, not zero.  I also considered that the sum is geometric, but I do not believe that is the case, because both $a_k$ and $k$ are variables that are not constant. I was thinking that I could argue if the series $a_k$ converges to a point, eventually the terms of the sequence will be so close to each other that it resembles a constant multiplier, but that argument is not very strong.  Any tips, hints, or leads at the solution are appreciated!
17618
I have this book Calculus, Ninth Edition by Varberg, Purcell, and Rigdon; there's a particular point of a theorem (and another line after that) about Infinite Series that I really don't understand. I quote from Section 9.7, Operations on Power Series:     Let $$f(x) = \sum_{n=0}^{\infty} a_n \cdot x^n$$ and $$g(x) =\sum_{n=0}^{\infty} b_n x^n$$, with both of these series converging at least for $|x| &lt; r$.    If the operations of addition, subtraction, and multiplication are performed on these series as if they were polynomials, the resulting series will converge for $|x| &lt; r$ and represent $f(x) + g(x)$, $f(x) - g(x)$, and $f(x) \cdot g(x)$, respectively. If $b_0 \neq 0$  the corresponding result holds for division, but we can guarantee its validity only for $|x|$ sufficiently small....We mention that the operation of substituting one power series into another is also legitimate for $|x|$ sufficiently small, provided that the constant term of the substituted series is zero.   What does it mean by "sufficiently small"? How do I find the limit of how large $|x|$ can be before divisions and substitutions between power series become invalid? My book doesn't go into detail on the exact nature of this restriction and doesn't include a proof of the theorem. If I were dividing one power series by another I don't really know what I should be looking for. I wasn't able to find information about convergence sets for a series generated by two power series divided by each other.
88868
I am looking at the series   $$ \sum_{n=1}^\infty\frac{(-1)^n}{n}.$$  This series converges (conditionally) by the alternating series test.  How can I compute its limit, which is equal to -log(2)?  a) I considered $I_n = \int_{0}^1$ $\frac{x^n}{1+x} dx$ -- and showed that this goes to 0, as n goes to infinity (use dominated convergence theorem).  b) I computed [$I_k$ + $I_{k-1}$] (for k $\ge$ 1), and it equals $\frac{1}{k}$.  I should be using my results from a) and b) to compute the above limit, which is part c) of the question that I am working on.  So, I have $$ \sum_{n=1}^\infty\frac{(-1)^n}{n}  = \sum_{n=1}^\infty(-1)^n[I_n+I_{n-1}].$$    Now I'm stuck.  How should I proceed from here?    I checked, by computing explicitly, that $-I_0$ = -log(2).  But I don't know how to arrive at $-I_0$ from the summation that I'm currently at.  Thanks in advance,
54998
While trying some problems along with my friends we had difficulty in this question.   True or False: The value of the infinite product $$\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\biggr)$$ is $1$.   I couldn't do it and my friend justified it by saying that since the terms in the product have values less than $1$, so the value of the product can never be $1$. I don't know whether this justification is correct or not. But i referred to Tom Apostol's Mathematical Analysis book and found a theorem which states, that    The infinite product $\prod(1-a_{n})$ converges if the series $\sum a_{n}$ converges.   This assures that the above product converges. Could anyone help me in finding out where it converges to? And,   Does there exist a function $f$ in $\mathbb{N}$ ( like $n^{2}$, $n^{3}$) such that $\displaystyle \prod\limits_{n=1}^{\infty} \Bigl(1-\frac{1}{f(n)}\Bigr)$ has the value $1$?
72585
Suppose $f$ is some real function with the above property, i.e.  if $\sum\limits_{n = 0}^\infty {x_n}$ converges, then $\sum\limits_{n = 0}^\infty {f(x_n)}$ also converges.   My question is: can anything interesting be said regarding the behavior of such a function close to $0$, other than the fact that $f(0)=0$?
34945
I am sorry that my question is not whether $\sum^\infty_{n=2} \frac{1}{\ln{(n!)}}$ convergent or not.  My question is how to compute the sum of series:  $$\sum^\infty_{n=2} \frac{1}{\ln{(n!)}},$$ It is a question in a textbook, and I have worked on it for days, but fail.
89687
Does $\displaystyle\sum^\infty_{n=1} \ln(\frac {2^n+1}{2^n})=\sum^\infty_{n=1} \ln(1+\frac 1 {2^n})$ converge?   Condensation, root, ratio and limit comparison tests don't help, I don't know what to do... Any hints please?  Note: no integral test.
60890
I have the following problem: $$\sum_{k=1}^{\infty}\frac{2^{k}k!}{k^{k}}$$  I only need to find whether the series converges or diverges. My initial thinking was to use the ratio test. I hit a stump after simplifying my limit, though:  $$\lim_{k}\frac{2k^{k}(k+1)}{(k+1)^{k+1}}$$  Normally I'd just derive by L'Hopital's Rule, but the derivative is pretty long and complicated. I doubt that's the correct route.
61800
Determine this series converge or diverge, and if it converges, find its value.  $$\sum_{n=1}^{\infty}\frac{\tan{n}}{2^{n}}$$  This was too hard for me, as unboundedness of $\tan{x}$ at infinite number of poles make harder to guess the behavior of value of tangent function at positive integers.  Thanks.
16813
I would like to know: How come that  $$\sum_{n=1}^\infty n x^n=\frac{x}{(x-1)^2}$$  Why isn't it infinity?
107026
$\lim_{x\to\infty} a_n$ defined as $a_{n+1}=a_n+\frac{1}{a_n}$ with $a_1=1$.  I've tried writing out the first few terms and this sum is increasing, but progressively less-so over time. I'm just not certain how to go about caclulating the limit here. I feel that it does have a finite limit, but I'm not certain of what it is.
103880
Just checking. $2^n$ ($n \to \infty$) tends to $\infty$. + $3^n$ also ($n \to \infty$) tends to $\infty$  so the sum gets me $\infty$.  Now $(\infty)^{1/\infty}$ : $(\infty)^0 = 1$  I see no other way.  Theorem: $n^{\frac{1}{n}} = 1$ as $n \to \infty$.  Analogous:  $(2^n - 3^n)^\frac{1}{n}$ = $(\infty  - \infty)^0$ = 1 ???  Is it ok to suppose infinity on any integer $k^n$ as $n$ goes to infinity ?
28180
Does the following sequence converge or diverge? If it converges, find its limit.  $\frac{1^1}{n^2}+\frac{2^1}{n^2}+........+\frac{n^1}{n^2}$  I tried to get a general formula for this sequence and simply got:  $\frac{n}{n^2}$ or $\frac{1}{n}$  This led me to think that the sequence converges to $0$ since the limit of this function is $0$. But this limit is incorrect. This tells me the general formula I got for the sequence is incorrect. But I am not sure why.  Any help?
10995
How do I find:  $$\prod_{n=1}^\infty\; \left(1+ \frac{1}{\pi ^2n^2}\right) \quad$$  I am pretty sure that the infinite product converges, but if it doesn't please let me know if I have made an error. Also, could I have a nice explanation as to how would someone arrive to the answer.  Thanks alot.