All Questions
111
questions
0
votes
1
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52
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How to prove the sum of limits theorem for a finite N number of limits? [duplicate]
I was reading a book with sequences and it proved that given two sequences $A$ and $B$ which both converge, then $\lim(A+B) =\lim(A)+\lim(B)$.
However, the sum of $N$ limits $$\lim(A_1+A_2+A_3+\dots)=\...
0
votes
0
answers
45
views
Is this a sufficient condition to interchange infinite sums?
I came across this wikipedia article, which has the following result:
Theorem 5: If $a_{m,n}$ is a sequence and $\lim_{n \to \infty} a_{m,n}$ exists uniformly in $m$, and $\lim_{m \to \infty} a_{m,n}$...
4
votes
1
answer
91
views
Why $\infty=\sum_{i=1}^\infty \frac{1}{n+i}\neq\lim_{n\rightarrow\infty}\sum_{i=1}^n \frac{1}{n+i}=\log2$?
I was wondering why $\sum_{i=1}^\infty \frac{1}{n+i}$ diverges but $\lim_{n\rightarrow\infty}\sum_{i=1}^n \frac{1}{n+i}=\log2$. While assuming integral as limit of series, we find out that:
$$
\int_1^...
2
votes
4
answers
158
views
A problem on finding the limit of the sum
$$u_{n} = \frac{1}{1\cdot n} + \frac{1}{2\cdot(n-1)} + \frac{1}{3\cdot(n-2)} + \dots + \frac{1}{n\cdot1}.$$
Show that, $\lim_{n\rightarrow\infty} u_n = 0$.
The only approach I can see is either ...
0
votes
4
answers
196
views
How to evaluate $\sum\limits_{n=3}^ \infty \frac{1}{n \ln(n)(\ln(\ln(n)))^2}$
I saw this problem : Prove that $\sum\limits_{n=3}^ \infty \frac{1}{n \ln(n)(\ln(\ln(n)))^2}$ converges, this is an easy problem could be proved using Cauchy condensation test twice.
$$\sum_{n=3}^ \...
5
votes
1
answer
183
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How to rigorously prove that $\sum\limits_{n=1}^ \infty( \frac{1}{4n-1} - \frac{1}{4n} )=\frac{\ln(64)- \pi}{8}$?
How to rigorously prove that $\sum\limits_{n=1}^ \infty\left( \frac{1}{4n-1} - \frac{1}{4n}\right) =\frac{\ln(64)- \pi}{8}$ ?
My attempt
$$f_N(x):= \sum_{n=1}^ N \left(\frac{x^{4n-1}}{4n-1} - \frac{x^...
2
votes
4
answers
273
views
How did Rudin change the order of the double sum $\sum_{n=0}^\infty c_n\sum_{m=0}^n\binom nma^{n-m}(x-a)^m$?
I see many people change the order of sum but I don't understand how they did that.
Is there is a way to change the order of the sum, $\sum\limits_{k=a}^n\sum\limits_{j=b}^m X_{j,k}$ and $\sum\...
3
votes
2
answers
296
views
if $\lim\limits_{n \to \infty} b_n =0 $ then how to prove that $\lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k}=0$
in Problems in Mathematical Analysis I problem 2.3.16 a),
if $\lim\limits_{n \to \infty}a_n =a$, then find $\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}$
The proof that ...
1
vote
1
answer
230
views
Compute $\lim\limits_{n\rightarrow+\infty}(\sum\limits_{i=1}^n(1+\frac{i}{n})^i)^{\frac{1}{n}}$
Here is a question in calculus. Compute the limit of the sequence: $\lim\limits_{n\rightarrow+\infty}(\sum\limits_{i=1}^n(1+\frac{i}{n})^i)^{\frac{1}{n}}$?
There are in general three ways to compute ...
4
votes
1
answer
89
views
A conjecture involving series with zeta function
Recently, I tried to evaluate a limit proposed by MSE user Black Emperor. In the process of evaluating the limit, I have obtained the following equality.
$$
\lim_{N\rightarrow \infty} \sum_{n=0}^{N-2}{...
6
votes
3
answers
160
views
Does $\lim_{k\to\infty}\sum_{n\ge1}\left(\frac{n!}{n^n}\right)^k$ converges to 1?
Context:
I was randomly putting infinite sums of this form ($k\in\mathbb N)$ in online calculators. $$\sum_{n\ge1}\left(\frac{n!}{n^n}\right)^k$$
I noticed that as $k$ increases, the sums becomes more ...
0
votes
1
answer
104
views
Asymptotic behavior of a sequence
I'm trying to generalize my previous question. Suppose that function $f:\mathbb{N}\to \mathbb{R}$ satisfies the following conditions:
$1. \ \ \forall i\in \mathbb{N}:f(i) = 2^{-l_i} \text{ for some} \ ...
2
votes
3
answers
336
views
Lower bound for a sequence
Following my previous question, suppose that $l_1\le l_2\le \dots $ are natural numbers such that $$\sum_{i=1}^{+\infty}2^{-l_i}\le1$$I want to find a lower bound on $l_i$ in order to compute the ...
5
votes
4
answers
342
views
Limit of the difference between two sequences
Suppose that $l_1, l_2,\dots$ are natural numbers such that $$\sum_{i=1}^{+\infty}2^{-l_i}\le1 \tag{1}$$ and let $$a_n=\frac{1}{n}\sum_{i=1}^{n}l_i - \log_2n \tag{2}$$ I think $\lim_{n \to \infty} a_n$...
1
vote
2
answers
70
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Find $\lim\limits_{n\rightarrow\infty}\sqrt[n]{\sum_{k=0}^{n} \frac{(-1)^k}{k+1}\cdot2^{n-k}\cdot\binom{n }{k}}$
Find
$$\lim_{n\rightarrow\infty}\sqrt[n]{\sum\limits_{k=0}^{n} \frac{(-1)^k}{k+1}\cdot2^{n-k}\cdot\binom{n }{k}}$$
My attemt: By the binomia theorem, we have
$$(1-x)^n={\sum_{k=0}^{n} \binom{n }{k} (...