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213 votes
5 answers
45k views

The sum of an uncountable number of positive numbers

Claim: If $(x_\alpha)_{\alpha\in A}$ is a collection of real numbers $x_\alpha\in [0,\infty]$ such that $\sum_{\alpha\in A}x_\alpha<\infty$, then $x_\alpha=0$ for all but at most countably many $\...
Benji's user avatar
  • 5,910
115 votes
25 answers
17k views

Can an infinite sum of irrational numbers be rational?

Let $S = \sum_ {k=1}^\infty a_k $ where each $a_k$ is positive and irrational. Is it possible for $S$ to be rational, considering the additional restriction that none of the $a_k$'s is a linear ...
User1's user avatar
  • 1,841
61 votes
3 answers
3k views

Proving $\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots\right)$

The equality$$\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots\right)\tag{1}$$follows from the fact that the sum of the first series ...
José Carlos Santos's user avatar
58 votes
7 answers
25k views

Is it possible to write a sum as an integral to solve it?

I was wondering, for example, Can: $$ \sum_{n=1}^{\infty} \frac{1}{(3n-1)(3n+2)}$$ Be written as an Integral? To solve it. I am NOT talking about a method for using tricks with integrals. But ...
Amad27's user avatar
  • 11.2k
42 votes
6 answers
7k views

If $f(x)=\frac{1}{x^2+x+1}$, how to find $f^{(36)} (0)$?

If $f(x)=\frac{1}{x^2+x+1}$, find $f^{(36)} (0)$. So far I have tried letting $a=x^2+x+1$ and then finding the first several derivatives to see if some terms would disappear because the third ...
chrismc's user avatar
  • 621
39 votes
12 answers
90k views

Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? [duplicate]

Can you please explain why $$ \sum_{k=1}^{\infty} \dfrac{k}{2^k} = \dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots = 2 $$ I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{...
jeebee's user avatar
  • 433
37 votes
3 answers
3k views

An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$

$n$ is a positive integer, then $$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53.$$ please don't refer to the famous $1+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$. I want to find a ...
ziang chen's user avatar
  • 7,819
36 votes
5 answers
9k views

use of $\sum $ for uncountable indexing set

I was wondering whether it makes sense to use the $\sum $ notation for uncountable indexing sets. For example it seems to me it would not make sense to say $$ \sum_{a \in A} a \quad \text{where A is ...
Beltrame's user avatar
  • 3,116
34 votes
3 answers
6k views

Why does Wolframalpha think that this sum converges?

Looking at the sum: $$\sum_{n=1}^\infty\tan\left(\frac\pi{2^n}\right)$$ I'd say that it does not converge, because for $n=1$ the tangent $\tan\left(\frac\pi 2\right)$ should be undefined. But ...
imanoob's user avatar
  • 555
32 votes
1 answer
1k views

Geometric representation of Euler-Maclaurin Summation Formula

In Tom Apostol's expository article (here's a free link), upon seeing the figure below (or this from the Wolfram project) I was expecting more diagrams to come to continue the error decomposition of ...
Lee David Chung Lin's user avatar
31 votes
1 answer
1k views

Prove a strong inequality $\sum_{k=1}^n\frac{k}{a_1+a_2+\cdots+a_k}\le\left(2-\frac{7\ln 2}{8\ln n}\right)\sum_{k=1}^n\frac 1{a_k}$

For $a_i>0$ ($i=1,2,\dots,n$), $n\ge 3$, prove that $$\sum_{k=1}^n\frac{k}{a_1+a_2+\cdots+a_k}\le\left(2\color{red}{-\frac{7\ln 2}{8\ln n}}\right)\sum_{k=1}^n\frac 1{a_k}.$$ The case without $\...
Tianlalu's user avatar
  • 5,217
29 votes
2 answers
829 views

How to prove $\sum_{n=0}^{\infty} \frac{1}{1+n^2} = \frac{\pi+1}{2}+\frac{\pi}{e^{2\pi}-1}$

How can we prove the following $$\sum_{n=0}^{\infty} \dfrac{1}{1+n^2} = \dfrac{\pi+1}{2}+\dfrac{\pi}{e^{2\pi}-1}$$ I tried using partial fraction and the famous result $$\sum_{n=0}^{\infty} \...
user avatar
29 votes
1 answer
570 views

A nice Combinatorial Identity

I am trying to show that $\forall N\in\mathbb{N}$, $$\sum\limits_{n=0}^{N}\sum\limits_{k=0}^{N}\frac{\left(-1\right)^{n+k}}{n+k+1}{N\choose n}{N\choose k}{N+n\choose n}{N+k\choose k}=\frac{1}{2N+1}$$ ...
Harmonic Sun's user avatar
  • 1,984
28 votes
4 answers
23k views

Summation Symbol: Changing the Order

I have some questions regarding the order of the summation signs (I have tried things out and also read the wikipedia page, nevertheless some questions remained unanswered): Original 1. wikipedia ...
TestGuest's user avatar
  • 1,053
25 votes
3 answers
6k views

Does uncountable summation, with a finite sum, ever occur in mathematics?

Obviously, “most” of the terms must cancel out with opposite algebraic sign. You can contrive examples such as the sum of the members of R being 0, but does an uncountable sum, with a finite sum, ...
Mike Jones's user avatar
  • 4,470

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