Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [summation]

Questions about evaluating summations, especially finite summations. For infinite series, please consider the (sequences-and-series) tag instead.

The notation $\sum\limits_{i=1}^na_i$ means $a_1+\ldots +a_n$.

Use for sums of infinite series and questions of convergence; use for questions about finite sums and simplification of expressions involving sums.

17770 questions
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A sum of powers formula? (I was just playing around, I don't know if this is correct)

So for reference, I'm just a high school student, who doesn't know that much maths stuff. But I noticed a pattern in binary numbers where, if you have all ones that is one less than if you have the next value (e.g. $1111$ is $1$ less than $10000$ -…
Josh
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Proving that $\sum_{k=0}^{\infty} \frac{2^k}{{2k \choose k}}=2+\frac{\pi}{2}$

Using tha integral representation of the reciprocal of the binomial co-efficient: $${n \choose j}^{-1}=(n+1)\int_{0}^{1} x^j (1-x)^{n-j}~ dx,~~~~(1)$$ here, we state and prove an interesting sum that $$\sum_{k=0}^{\infty} \frac{2^k}{{2k \choose…
Z Ahmed
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Sum $\sum_{n=1}^{\infty}\frac{1}{n^3}$

Probably you would have known the answer of $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.$ I then became curious about the reciprocal of cubes, not squares, like $$\sum_{n=1}^{\infty}\frac{1}{n^3}$$Can you find any answer to this? I have no…
user
  • 551
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Algebraic Way to Prove $\sum_{r=0}^{n} \frac{(2n)!}{((n-r)!)^2(r!)^2} = \binom{2n}{n}^2$

$\sum_{r=0}^{n} \frac{(2n)!}{((n-r)!)^2(r!)^2} = \binom{2n}{n}^2$ This identity can be easily proven by mathematical induction. However, is there any algebraic way to sum it up to $\binom{2n}{n}^2$?
Liew Jia Hao
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Solving for a variable in a summation?

This semester, I'm taking Calculus two. I'm working on a project for myself where I will need to solve the for a variable, inside a summation, such as "i" in the equation the one below: $$ 0=-20+\sum _{n=1}^{\infty…
findingmyway
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Problem with summation by method of difference

Question: What would be the result of: $$\sum_{k=1}^{n}\frac{1}{n(n+2)}$$ My Approach: Let $T_n$ denote the $n^{th}$ term of the given series. Then we have $$T_1=\frac12 \left(\frac11-\frac13\right)$$ $$T_2=\frac12…
General Kenobi
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Question about finding the value of an infinite sum

What is the value of: $$\sum_{k=0}^\infty \frac{1}{(4k+1)^2}?$$ I realised that $$\sum_{n=2,4,6,8,...} \frac{1}{n^2} + \sum_{n=1,3,5,7,...} \frac{1}{n^2} = \sum_{n \geq 1 } \frac{1}{n^2}$$ $$\sum_{n \geq 1 } \frac{1}{4n^2}+\sum_{n \geq 0 }…
Darius Chitu
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Do not understand what is meant by "in terms of n"

I took calculus 5 years ago so I am unaware of what is meant by in terms of the variables in this summation example. specifically these two questions. please help me understand what the steps are to solving these. I was unable to google my…
John Adair
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Evaluating Summation in Iverson's Brackets

I am trying to find some reasonable upper bound on $$\sum_k [0 < k\alpha \leq n] [\{k\alpha\} < 1/k]$$ where $\alpha$ is irrational, the brackets are Iverson's notation ($1$ if true, $0$ if false), and $\{x\} = x - \lfloor x \rfloor $. If we select…
lamlame
  • 349
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Where does the -2 +1 come from in this sigma notation?

I am solving this equation: My issue arises on line 2 where we have (n + 1 - 2 + 1)(n + 1 + 2)/2. This is what i understand. We have the formula n! = n(n+1)/2. Subbing values into the equation yields us with 10[(n+1)(n+1 + 1)/2] however we need to…
archhmod
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how to change where the limits start

$$ \begin{aligned} \sum_{n=1}^{N} \frac{1}{n^{2}} & \leq 1+\sum_{n=2}^{N} \frac{1}{n^{2}-n} \\ &=1+\sum_{j=1}^{N-1} \frac{1}{n(n+1)} \end{aligned} $$ In this summation, the person changes the summation of expression $\frac{1}{n^{2}-n}$ to…
Joseph Rock
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A question about meaning of a notation

When we have a sum of the form $$ \sum_{cyc} \dfrac{ (a+b)(a+c) - bc }{(b-c)(b^3-c^3)} $$ Does this mean: $$ \dfrac{ (a+b)(a+c) - bc }{(b-c)(b^3-c^3)} + \dfrac{ (b+c)(b+a) - ac }{(a-c)(a^3-c^3)} + \dfrac{ (c+b)(c+a) - ba }{(b-a)(b^3-a^3)} $$ ?
James
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How to solve $11+192+1993+19994+...+1999999999$

I thought of converting this to the $\sum$ notation and then simplifying it further. On simplifying, it turned out to be a combination of two summation notations. Here's what I got $$\sum_{n = 1}^{9} \left[ 10^n + 90 \left( \sum_{a = 0}^{n - 2} …
Rajdeep Sindhu
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Why does $\sum^N_{n=0}\frac12\leq\sum^N_{n=0}\sum^{2^{n+1}-1}_{r=2^n}\frac1r\leq\sum^N_{n=0}1=\frac{N+1}2\leq\sum^{2^{N+1}-1}_{r=1}\frac1r\leq N+1$?

Consider the following inequalities: $$\frac{1}{2} \leq \sum^{2^{n+1}-1}_{r=2^n}\frac{1}{r}\leq 1 \tag{1}$$ Upon summing over $(1)$ from $n=0$ to $n=N$, we obtain $$\sum^N_{n=0}\frac{1}{2}\leq…
Chern-Simons
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change of index on -infinite and +infinite summation

I am simplifying the following equation. $$ =\sum_{k=-\infty}^{\infty}{ x_k \sum_{n=-\infty}^{\infty}{h_{n-k}}}$$ If sum index for $n$ is changed to $r=n-k$, where $n = r+k$. $$ =\sum_{k=-\infty}^{\infty}x_k…
endamusiyen
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