Let $p:\mathbb{C}\rightarrow \mathbb{C}$, $z\mapsto a_kz^k+a_{k-1}z^{k-1}+\ldots +a_1z+a_0$ be a polynomial with coefficients $a_j\in \mathbb{C}$, $j=0, \ldots ,k$ and $a_k\neq 0$. For which $q\in \mathbb{C}$ is the series $\displaystyle{\sum_n p(n)q^n}$ convergent? We have that \begin{align*}\sum_n p(n)q^n&amp;=\sum_n \left (a_kn^k+a_{k-1}n^{k-1}+\ldots +a_1n+a_0\right )q^n\\ &amp; =a_k \sum_n n^k q^n +a_{k-1} \sum_n n^{k-1} q^n +\ldots +a_1 \sum_n n q^n + a_0 \sum_n q^n \end{align*} So we have to check the convergence of a series of the form $\displaystyle{\sum_n n^jq^n}$. Let $a_n=n^jq^n$. Then from the ratio test we have that $$\left |\frac{a_{n+1}}{a_n}\right |=\left |\frac{(n+1)^jq^{n+1}}{n^jq^n}\right |=\left |\left (1+\frac{1}{n}\right )^jq\right |$$ To get convergence we want that this is smaller than $1$, right? So we set $\left |\left (1+\frac{1}{n}\right )^jq\right |&lt;1$ and solve for $q$ ?
120711
Let $p(x)$ be a polynomial of the real variable $x$ of degree $k\geq 1$ .Consider the power series $$f(z)=\sum_{n=0}^{\infty}p(n)z^n$$ where z is a complex variable .Then the radius of convergence of $f(z)$ is ?
376
Let $a_n$ be a positive real sequence such that the series $\sum a_n$ converges.  I was asked to prove that under such circumstances $\sum a_n^{\frac{n}{n+1}}$ converges.  The previous sum can be rewritten as $\sum \frac{a_n}{a_n^{1/(n+1)}}$.  I can't prove the convergence of this last series... I'm sure comparison test is enough here, but other than $a_n \rightarrow 0$ there is no information that helps.  Thanks for your help.
1983
Let $a_1,a_2,a_3,\ldots$ be reals. Prove that if $\sum_{n=1}^{\infty} a_n^{2}$ converges, then so does $\sum_{n=1}^{\infty} \frac {a_n}{n}$.   For this I have shown the case for when $ a_n^{2} \le\frac {|a_n|}{n}$ $\Rightarrow$ $ |a_n|\le\frac {1}{n}$ $\Rightarrow$ $\frac {|a_n|}{n} \le \frac{1}{n^{2}}$ and we know that $\sum_{n=1}^{\infty} \frac {1}{n^{2}}$ converges and hence $\sum_{n=1}^{\infty}\frac {a_n}{n}$ converges by the comparison test. Now considering $ a_n^{2} \ge\frac {|a_n|}{n}$ $\Rightarrow$ $\frac {|a_n|}{n} \le a_n^{2}$ $\rightarrow$ combining the two cases for any n we have: $\frac {|a_n|}{n}\le\frac{1}{n^{2}}+a_n^{2}$ Hence using the comparion test again we know that $\sum_{n=1}^{\infty} a_n^{2}$ converges and $\sum_{n=1}^{\infty} \frac {1}{n^{2}}$ converges hence the sum converges so we can conclude that $\sum_{n=1}^{\infty} \frac {a_n}{n}$ is absoluetly convergent $\Rightarrow$ convergent.  Not to sure if this is correct, any help would be much appreciated, many thanks.
108215
Studying for an exit exam and it's been years since I work with any series expansions. Here's a past problem:  Show that $$\sum_{n=1}^{\infty} \frac{z^n}{1+z^{2n}}$$ converges to an analytic function on the set $A=\{z \in \mathbb{C}: |z|&lt;1 \}$.  I'm looking to use Taylor's theorem for this problem but how would I compute the radius of convergence? Do I have to do a comparison test?
104458
Suppose that $$Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} $$and $$P(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}.$$ How do I find $$\lim_{x\rightarrow\infty}\frac{Q(x)}{P(x)}$$ and what does the sequence $$\frac{Q(k)}{P(k)}$$ converge to?  For example, how would I find what the sequence $$\frac{8k^2+2k-100}{3k^2+2k+1}$$ converges to?  Or what is $$\lim_{x\rightarrow\infty}\frac{3x+5}{-2x+9}?$$   This is being asked in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: .
112684
$$ \sum_{n=2}^{\infty}\frac{1}{(\sqrt{n}+\sqrt{n+1})^p}\ln\left(\frac{n-1}{n+1}\right). $$ For which values of the parameter p is it convergent?
57722
Not sure how to approach this problem below. I tried assuming the conditional but that doesn't seem to get me anywhere.     If for a sequence of non-negative real numbers $\{a_n\}$, we have $$\sum{a_kb_k}$$ converges whenever $$\sum{b_k^2}$$ converges. then show that $$\sum{a_k^2}$$ converges
90705
Is there a sequence of polynomials $P_n$ such that   $\displaystyle\lim_{n \to \infty} P_n(z) = \begin{cases} 1, &amp; \text{Im } z &gt; 0\\ 0, &amp; z \text{ is real,} \\ -1, &amp; \text{Im } z &lt; 0 \end{cases}$  I have no clue where to start from. Please provide some hints. Thanks in advance.
93786
Let $ a_{n} = \sum_{k=1}^{n} \frac{n}{n^{2}+k}$ . I would like to know whether the given sequence converges.   I see  that,  $ a_{n} = \sum_{k=1}^{n} \frac{n}{n^{2}+k}= \sum_{k=1}^{n} \frac{1}{n+\frac{k}{n}}.$ When $n$ gets sufficiently large the contribution by the $ \frac{k}{n} $ term is diminishing and $ a_{n} &lt; \sum_{k=1}^{n} \frac{1}{n} = 1 $.   Thank you.
11017
Let $p(z)=z^{n}+a_{n-1}z^{n-1}+...+a_{0}$ be a polynomial of degree $n\geq1$.  How can you prove that either $p(z)=z^{n}$ or there exists $z'$ with $|z'|=1$ such that $|p(z')|&gt;1$.  Maybe we can use the maximum modulus principle and consider $q(z)=z^{n}p(1/z)$.
26570
The last problem I posted had a wrong statement. I recovered the correct one.  Let $(a_n)$ be a sequence of positive real numbers. Prove that $\sum_{n\geq 0}a_n$ converges iff $\displaystyle \sum_{n\geq 0} \frac{a_n}{\sum_{k=0}^n a_k}$ converges  The direct statement is easy to prove using comparison test. I'm (again!) stuck with the converse. I tried summation by part, without success. Note that the convergence of $\displaystyle \sum_{n\geq 0} \frac{a_n}{\sum_{k=0}^n a_k}$ implies $\displaystyle \frac{a_n}{\sum_{k=0}^n a_k} \to 0$ I don't know what to do next ...
81852
This question is in relation to the space $C(I)$, $I = [a, b]$. Define $P_n =\{ a_0+\dots+a_nx^n \mid a_i \in \mathbb{R}\}$ (any or all $a_i$ could be zero); clearly $P_n \subset C(I)$. The norm I'm using is $\lVert f\rVert_I = \sup_I |f(x)|$. Is $P_n$ closed under $\lVert\cdot\rVert_I$?   I am almost sure the answer is "yes", but I can't seem to prove it. My first instinct was to biject  $P_n$ to $\mathbb{R}^{n+1}$, using coefficients as coordinates, and prove that sequences of degree-$n$ polynomials converge to degree-$n$ polynomials, but I can't prove that the metric $\lVert\cdot\rVert_I$ is equivalent to the standard metric on $\mathbb{R}^{n+1}$. Next, intuitively given a function $f \in C(I), \notin P_n$ I should be able to find some $\epsilon &gt; 0$ such that there are no low-degree polynomials "nearby", then that function was not a limit point of $P_n$ so $P_n$ is closed. Again I have no idea how to prove this. What should I do?
52961
Does the sequence $$\displaystyle \frac{n}{\sum\limits_{k=1}^{n}\Big(\frac{k}{k+1}\Big)^k}$$   converge?   Attempt. Since $\Big(\frac{k}{k+1}\Big)^k \rightarrow 1/e\neq 0$ and the terms are positive, the series $\sum\limits_{k=1}^{\infty}\Big(\frac{k}{k+1}\Big)^k$ diverges to $+\infty$. I find hard to determine if $n$ or the sum goes faster to $+\infty.$   Thanks in advance.
68839
I want to prove the following:  Let $\{ a_n \}$ and $\{ b_n \}$ be two sequences with $a_n&gt;0$ and $b_n&gt;0$ for all $n \geq N$, and let $$c_n = b_n - \frac{a_{n+1}b_{n+1}}{a_n}$$  Then   If there exists $r&gt;0$ such that $0&lt;r\leq c_n \text{ ; } \forall n\geq N$ then $\displaystyle\sum a_n$ converges. If $c_n\leq0$ for $n\geq N$ and if $\displaystyle\sum \dfrac{1}{b_n}$ diverges, then so does $\displaystyle\sum a_n$.   So far I know how to use and prove Cauchy's, D'Alambert's and the integral criterion, so I'd like a hint on either using those or a new idea. The book suggests that for $1$, I show that  $$\sum\limits_{k = N}^n {{a_k} \leq  \frac{{{a_N}{b_N}}}{r}} $$  and for $2$, to prove that $\displaystyle\sum a_n$ dominates $\displaystyle\sum \dfrac{1}{b_n}$  I'd like to prove this with previous theory on series convergence (Cauchy's and/or D'Alambert's preferrably, comparison test) since it is what precedes the problem, and would appreaciate great HINTS rather than answers.   I can't seem to understand the inequalities. I mean, Cauchy's and D'Alambert's criteria reveal that the proof relies on the fact that a geometric series of with ratio $|r| &lt; 1$ converges, but I can't understand the motivation in this proof.    Summing up the work.  For 1:  $$\eqalign{   &amp; {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{b_{N + 1}}  \cr    &amp; {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{b_{N + 2}}  \cr    &amp; {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{c_{N + 2}} + \frac{{{a_{N + 3}}}}{{{a_N}}}{b_{N + 3}} \cr} $$  Induction over $n$ we get  $${b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{c_{N + 2}} +  \cdots  + \frac{{{a_{N + n}}}}{{{a_N}}}{c_{N + n}} + \frac{{{a_{N + n + 1}}}}{{{a_N}}}{b_{N + n + 1}}$$  So  $$\eqalign{   &amp; {b_N} \geqslant r\left( {1 + \frac{{{a_{N + 1}}}}{{{a_N}}} + \frac{{{a_{N + 2}}}}{{{a_N}}} +  \cdots  + \frac{{{a_{N + n}}}}{{{a_N}}}} \right) + \frac{{{a_{N + n + 1}}}}{{{a_N}}}{b_{N + n + 1}}  \cr    &amp; {b_N} &gt; r\left( {1 + \frac{{{a_{N + 1}}}}{{{a_N}}} + \frac{{{a_{N + 2}}}}{{{a_N}}} +  \cdots  + \frac{{{a_{N + n}}}}{{{a_N}}}} \right)  \cr    &amp; \frac{{{a_N}{b_N}}}{r} &gt; {a_N} + {a_{N + 1}} + {a_{N + 2}} +  \cdots  + {a_{N + n}} = \sum\limits_{k = N}^n {{a_k}}  \cr} $$  As desired.
46773
When does $P(x)=\sum_{n=0}^{\infty}{\frac{n!}{n^n}\cdot x^n}$ with $x\in \mathbb{C}$ converge?   My Attempt:$$\bigg |\frac{a_n}{a_{n+1}}\bigg |=\left(1+\frac{1}{n}\right)^n\to e$$ So it converges for $|x|&lt;e$. Now I wanted to check wheter the series converges for $e$ or $-e$.  I fail to either prove the divergence or convergence. Can you help?
105027
I have the following question regarding Modes of Convergence          Recall that the Taylor series of the function $f(x) = \ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$. Let $f_{N}(x)=\sum_{n=1}^{N}(-1)^{n-1}\frac{x^n}{n}$ be the partial sum. Determine whether $f_N$ converges pointwise, uniformly, or in the $L^2$ sense to $f$ on $-1\leq x\leq1$.      I am kinda sketchy on whether it converges pointwise - well in general actually - for $-1$ since $\ln(1+-1) = \ln(0)$ is undefined. If I disregard this, let's take a look at what we know. $\ln(x+1)$ = $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$ converges on $|x|&lt;1$ by the ratio test. The ends would be of concern, but $1$ is okay. Not so much $-1$. Anywho, this is what I have.  Pointwise Converges: Consider $f(x) = \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$ and its $N$th partial sum $f_{N}(x)=\sum_{n=1}^{N}(-1)^{n-1}\frac{x^n}{n}$. The infinite series converges pointwise to $f(x)$ on $-1\leq x \leq 1$ if   $ \begin{eqnarray}  | \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}-\sum_{n=1}^{N}(-1)^{n-1}\frac{x^n}{n}|\\ |(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+...)-(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+...)|\\ |(x-x)+(-\frac{x^2}{2}+\frac{x^2}{2})+(\frac{x^3}{3}-\frac{x^3}{3})+(-\frac{x^4}{4}+\frac{x^4}{4})+(\frac{x^5}{5}-\frac{x^5}{5})+...| \longrightarrow 0\\ \end{eqnarray} $  as $N \rightarrow \infty$, for every $x\in[-1,1]$.  However, for some reasoning I am doubting it actually converges on the entire interval. I could just be over thinking it. The point is to show the error between $f(x)$ and $f_N(x)$ tends to zero as $N\rightarrow \infty$ for every $x$ on the interval and it seems to do just that.  Uniformly Convergence is similar; however, it focuses on uniform error between $f(x)$ and $S_N(x)$ tends to zero as $N\rightarrow \infty$. If want to conclude it also tends to zero. Same goes for $L^2$ sense.  Any suggestions? Thank you for your time and Thanks in advance for any feedback.
113957
Suppose that $f$ is an entire function such that $\forall \vert z \vert =R$, $f(z)\leq M$, then for each coefficient $C_k$ in its power series expansion $f(z)=\sum C_kz^k$, $$|C_k|\leq \frac{M}{R^k}$$  My approach was plugging in $f(Re^{i\theta})$ which is $\sum C_kR^ke^{ik\theta}$. So $$\sum C_kR^ke^{ik\theta} \leq M$$  and then suppose that $C_l = 0$ for $l\neq k$. Which leads to $$C_kR^ke^{ik\theta} \leq M$$ $$C_k \leq \frac{M}{R^Ke^{ik\theta}}$$  I am not sure what to do at this point or if my steps leading up to this are correct. Any guidance would be appreciated. Thanks