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Results tagged with sequences-and-series
Search options answers only
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user 1131
for questions about sequences and series, e.g. convergence, closed form expressions, etc. Note that there is a different tag for spectral sequences, and also note that MathOverflow is not for homework. Please consider consulting the online encyclopedia for integer sequences, if you are trying to identify a given sequence that you have found in your research.
6
votes
Two Equal Series?
Here is a link to a counterexample for conditionally convergent complex series.
- 61.9k
5
votes
Accepted
The sum of a series
I have no doubt that you have figured this exercise out by now but let me post the solution just for the fun of it.
Denote $U_k=\prod_{m=1}^k\frac 1{q^m-1}$ with the usual convention $U_0=1$.
We hav …
- 61.9k
10
votes
Accepted
If two functions are equal to their Newton series, is their composition also equal to its Ne...
Here goes, as promised.
Let $f$ be entire of order less than $1$, so $|f(z)|\le Ce^{|z|^p}$, $p<1$. Write the Newton polynomial
$$
P(x)=\sum_{k=0}^n\Delta^kf(0) {x \choose k}
$$
Note that $g(k)=f(k …
- 61.9k
3
votes
Approximate sum by an integral: valid or not?
The main observation:
Let $f,g$ be analytic in the disk $\{|z|\le 2\delta\}$ and real on the interval
$(-2\delta,2\delta)$. Assume that $f(0)=0$ and $f(x)<0$ for $0<|x|<2\delta$. Let $\psi$ be any $C …
- 61.9k
6
votes
Accepted
Pseudo-alternate series
That would be more than welcome on AoPS, College Playground. For MO, it is hardly appropriate.
The statement is always true. Start with the fact that if the sums $\sum_{i=k}^m b_i$ ($1\le k\le m\le …
- 61.9k
3
votes
Is there a theorem that says that there is always more than one way to "continue a finite se...
Normally, "valid" means "defined by a clear and short rule". Formally we may talk about having a generating program on some abstract machine that is much shorter than the sequence itself. The formal u …
- 61.9k
11
votes
Accepted
Given four conditionally convergent series, is there a single sequence of naturals such that...
There is a counterexample with 4 series. Notice that the problem is equivalent to asking if for every sequence of vectors $X_j$ in $\mathbb R^4$ with lengths tending to $0$ and the infinite sum of abs …
- 61.9k
3
votes
Why is the following recurrent sequence convergent?
That is, indeed, more bark than bite, but the bark is somewhat louder than mlk presented it.
First of all, let us define $X_k=\max(|x_k|,|x_{k-1}|)$ and notice that ${2n+1\choose 2k+1}\ge 2^{\min(k,n- …
- 61.9k
6
votes
Proving convergence of solution of a fixed point equation
Convergence to $0$ is trivial: as you noticed yourself, the coefficient is just ${n-1\choose k,k,n-1-2k}$. Splitting the rest as $(\sqrt\alpha x)^k(\sqrt\alpha x)^k(1-x)^{n-1-2k}$, we can bound $f_n(x …
- 61.9k
123
votes
Accepted
Is $ \sum\limits_{n=0}^\infty x^n / \sqrt{n!} $ positive?
Looks like the computers really spoiled us :)
GH gave a perfectly valid answer already but the cheapest way to prove positivity is to write $\int_0^1(1-t^n)\log(\frac 1t)^{-3/2}\,\frac{dt}t=c\sqrt n …
- 61.9k
7
votes
Accepted
Coefficient bounds of an inequality
This is true.
I prefer to denote $q_i=p_i^{-1}$, $\alpha_i=a_ip_i$, $\beta_i=b_ip_i$, $A_i=\sum_{j=1}^i\alpha_j$, $B_i=\sum_{j=1}^i\beta_j$. Now we have to check that
$$
\sum_i q_i\beta_iA_i^k\le C\ …
- 61.9k
5
votes
Accepted
Convergence for a non-linear second order difference equation
I would suspect that non-classical arguments are needed to do so.
All you need to know is that $t\mapsto \frac 1{1+t}$ is a decreasing function, so for $0<x\le x'$ we have $\frac{\log(1+x')}{\log(1+x) …
- 61.9k
5
votes
Accepted
Asymptotics of Fourier coefficients of power-type functions
$$
\begin{aligned}
&\int_0^\pi t^\beta e^{iyt}\frac{dt}{t}dt=y^{-\beta}\int_0^{\pi y} t^{\beta} e^{it}\frac{dt}{t}=
\cr
&=y^{-\beta}e^{i\pi\beta/2}\left[\int_0^{\pi y} t^\beta e^{-t}\frac{dt}{t}+O(t …
- 61.9k
6
votes
Accepted
Ratio of Sequences Sum Inequality
Here is an old cheap trick that may be helpful. Assume that $\sum_i a_i=A$, $\sum_i b_i=B$ and $a_i\le b_if(b_i)$ where $f$ is a decreasing function tending to $0$. Choose $b$ so that $f(b)=\frac{A}{2 …
- 61.9k
10
votes
Accepted
The digit sum: $s(na)=s(nb)$
OK, by Seva's request I'm getting somewhat more serious :) Fix $a$, $b$. Take large $M$ to be chosen shortly. Take a $3MN$ digit number $n$ (something from $0$ to $10^{3MN}-1$, written with head zeroe …
- 61.9k