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.

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Probing a particular function

I've been playing with a particular function $$Q(n) = \sum_{i=1}^n i\cdot i!$$ in C++, and I'm trying to see if it is possible to find the following in an elegant way: 1) Is it possible to rewrite the above function in such a way that it admits a…
Millardo Peacecraft
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Replace $\sum$ with $\ln$

Reading an solution I get stuck at this step. $$ \frac{1}{2i} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \left( e^{2in} - e^{-2in} \right) \\ = - \frac{1}{2i} \left( \ln(1 + e^{2i}) - \ln(1- e^{-2i}) \right) $$ Could someone explain to me what is done…
iveqy
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closed form of summation of geometric theories

I am working on a discrete convolution problem. I am comparing the solution to what the solution manual has and it just doesn't make sense to me. what solution manual has: https://i.stack.imgur.com/ywkpg.jpg The parts in red box doesn't make sense…
D.Zou
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sum polylogarith convergence

I am working in the following series i get a result could you help to see the convergence any ideas $$\sum _{k=1}^{\infty } \frac{(-1)^{k+1} \text{Li}_k(-a)}{a}=\frac{a-\log (a+1)}{a^2}$$ where Lk is the polylogaritm.
user167276
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solve $\sum_{k=0}^a\binom{2a}{k}k$

solve $$\sum_{k=0}^a\binom{2a}{k}k$$ I solved $S=\sum_{k=0}^a\binom{2a}{k}$ using $\binom{2a}{k}=\binom{2a}{2a-k}$ and got $S=\frac{4^a+\binom{2a}{a}}{2}$ but this idea doesn't work with $$\sum_{k=0}^a\binom{2a}{k}k$$ is there any Hints or solution…
user130806
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How do I evaluate $\sum _{ i=1 }^{ 2^n } (\frac{i}{2^n} - \frac{i-1}{2^n})(1-\frac{i-1}{2^n})$

What does this sum equal ? $\sum _{ i=1 }^{ 2^n } (\frac{i}{2^n} - \frac{i-1}{2^n})(1-\frac{i-1}{2^n})$ The answer I'm getting is $-\frac{1}{2^{n+1}} - \frac{1}{2^{2n+1}} + \frac{1}{2^{n}}$ but I have a feeling this is wrong. I would appreciate…
user1068636
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Evaluating sums question $\sum_{k=2}^n \frac{1}{k^2-1}$

I need to use the identity $\frac{1}{k^2-1}=\frac{1}{2}(\frac{1}{k-1}-\frac{1}{k+1})$ to evaluate $\sum_{k=2}^n \frac{1}{k^2-1}$. I am confused about how to begin this proof. Also, how am I supposed to use the identity here? Do I need to use…
user45417
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what is $2\cdot 4\cdot 6\cdot 8 \cdot \ldots \cdot (2k+2)$?

I know that $2\cdot 4\cdot 6\cdot 8 \cdot \ldots \cdot (2k)$ is $2^kk!$ but what is the value of these terms up to the $(2k+2)^\text{th}$ term?
E Be
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Finding roots for nested summations

Hi I was wondering how do I Solve this question. I have to solve for the root. I can solve for it when there's one summation but it's nested. I'm not that good at solving summations, if I can get some guidance please. Question is related to…
user3276954
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summation of polynomial products

I need help in understanding how the summation of the product of two polynomials is written. $(a_{0} +a_{1}x +a_{2}x^{2})(b_{0} +b_{1}x + b_{2}x^{2}) =\\ (a_{0}b_{0})x^{0} + (a_{1}b_{0} + b_{1}a_{0})x + (a_{2}b_{0} +a_{1}b_{1} + a_{0}b_{2})x^{2} +…
Dan
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closed form for $\sum_{k=1}^n\left[ 2^k\binom{2n-k}{n-k}-2^{k+1}\binom{2n-k-1}{n-k-1}\right ]k^s$

$$\sum_{k=1}^n\left[ 2^k\binom{2n-k}{n-k}-2^{k+1}\binom{2n-k-1}{n-k-1}\right ]k^s$$ by some steps I got $$\sum_{k=1}^n\left[ 2^k\binom{2n-k}{n-k}-2^{k+1}\binom{2n-k-1}{n-k-1}\right ]k^s=\sum_{k=1}^n2^k\binom{2n-k-1}{n}\frac{k^{s+1}}{n-k}$$ and…
user130806
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Sum of the series of $\displaystyle \sum_{i=0}^ni^2 2^{n-1}$

Question: Evaluate $$ \sum_{i=0}^n i^2 2^{n-i}$$ In the previous questions on my question paper I have got: $$\sum_{i=1}^n a^i = \frac{a(1-a^n)}{1-a}$$ and $$\sum_{i=1}^nia^i=\frac{a(1-a^n)}{1-a}-na^{n+1}$$ My step for this evaluation is…
hlx98007
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Asymptotics of $\sum _{k=1}^n \sum _{j=1}^n \frac{j k}{j+k}$

I'm asked to find a simple asymptotical estimation of $\displaystyle \sum _{p=1}^n \sum _{q=1}^n \frac{p q}{p+q}$. I rewrote the sum as $\displaystyle \sum _{k=2}^{2 n}\sum_{p+q=k}\frac{pq}{p+q}= \sum _{k=2}^{2 n} \sum _{p=1}^{k-1} \frac{p…
Gabriel Romon
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Is it correct that $ \sum_{i=m}^na=(nm+1)a$

I study this in a book. Is it correct? Why? $$ \sum_{i=m}^na=(nm+1)a$$
Reza Mohammady
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Evaluation of $\sum_{n=1}^{\infty} \frac {x^{-n}}{n}$

I wanted to evaluate $$ \sum_{n=1}^{\infty} \frac {2^{-n}}{n} $$ And noticed that for any base it has a pattern, so now I want to know how to evaluate $$ \sum_{n=1}^{\infty} \frac {x^{-n}}{n} $$ I don't have any approach. The result is logarithmic.…
UserX
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