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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Solve $ \sum_{k = 1}^{ \infty} \frac{\sin 2k}{k}$

Solve $$ \sum_{k = 1}^{ \infty} \frac{\sin 2k}{k}$$ I first tried to use Eulers formula $$ \frac{1}{2i} \sum_{k = 1}^{ \infty} \frac{1}{k} \left( e^{2ik} - e^{-2ik} \right)$$ However to use the geometric formula here, I must subtract the $k=0$ term…
iveqy
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series of $\arctan \frac{2x^2}{n^2}$ .

Find : $$\sum_{n=1}^{\infty}\arctan \frac{2x^2}{n^2}.$$ Where $x\in \mathbb{R}$.
Tulip
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Finding a sum of alternating cubes

Find this sum, in terms of $n$: I have some hand-written hints from someone else, but can't read his writing:
VividD
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Solving this summation: $\sum_{i=1}^k i\cdot2^i$

$$\sum_{i=1}^k i\cdot2^i$$ I'm working on a recurrence relation question and now I'm stuck at this point. I have no idea how to simplify this down to something I can work with. Can I seperate the terms into $$\sum_{i=1}^k i \cdot \sum_{i=1}^k 2^i$$…
user136038
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Summation of $2^n$ where $n$ is even or odd

I need the summation formula for the following: $2^0+2^1+2^3+2^5+\cdots+2^n$, $n$ is even i.e. $2^0+2^2+2^4+2^6+\cdots+2^n$, $n$ is odd I am aware that the formula of $2^0+2^1+2^2+2^3+\cdots+2^n = 2^{n+1}-1$. But can anyone help me with a approach…
divyenduz
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Evaluating Summation of $5^{-n}$ from $n=4$ to infinity

The answer is $\frac1{500}$ but I don't understand why that is so. I am given the fact that the summation of $x^{n}$ from $n=0$ to infinity is $\frac1{1-x}$. So if that's the case then I have that $x=\frac15$ and plugging in the values I have…
jj103
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Summing $\sum_{n=0}^{\infty} n x^n$

I am trying to figure out how to do the infinite summation: $$ \sum_{n=0}^{\infty} n x^n \qquad 0 \leq x < 1$$ The series converges so it seems to me that the limit must exist, but I'm having difficulties trying to find an exact answer for it. It is…
Yellow
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Summation of numbers $ (n+1)(n+2)\cdots(n+m)$ over $n$

Verify that if $V_n = n(n+1)(n+2)\cdots(n+m)$ then $$V_{n+1} - V_n = (m+1)(n+1)(n+2)\cdots(n+m)$$ Given now that $U_n = (n+1)(n+2)\cdots(n+m)$ find sum of series $U_n$ from $N$ to $1$ in terms of $m$ and $N$. I have done the first part of the…
user140161
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Interesting combinatorial identities

Let $n$ be a strictly positive integer and let $j=0,\dots,n-1$. By using Mathematica I managed to guess the following identities: \begin{eqnarray} \sum\limits_{m=0}^{n-j-1} \binom{n-m-1}{j} \binom{n+m}{j} &=& \frac{1}{2} \binom{2 n}{2j + 1}…
Przemo
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summation of a series in which each term is product of nth term of two sequence

Is it possible to find the sum $\Sigma_{x=1}^n ((2x)(4x+1))$? If yes then can somebody please explain for me the formula?
Prerak
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$\sum_{k=1}^m n-k$ and $\sum_{k=m+1}^{2m} k$

i really can't find out why $$\sum_{k=1}^m n-k = -\frac12\left(m^2-2nm +m\right)$$ and why $$\sum_{k=m+1}^{2m} k = \frac12m\, (3m +1)$$ For the first one i really don't know where to start, but for the second one i tried to put the numbers one near…
Bman72
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How do we solve complicated summations

I am reading Introduction with Algorithms book and my first doubt arises in time analysis of my first insertions sort's algorithm uses these sigma problems. I learned about this in junior classes but never solved complicated ones.Even appendix in…
Abhimanyu Aryan
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How could I find the partial sum of this function?

$$\displaystyle\sum_{k = 0}^{n}\dfrac{k^2}{2^k}$$ What are the steps to finding the partial sum formula?
Waffle
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Prove: $\sum_{x=0}^{n} (-1)^x {n \choose x} = 0$

Is there a quick, fancy, way of proving sums such as this? Prove that: $$\sum_{x=0}^{n} (-1)^x {n \choose x} = 0$$ A recent homework assignment I turned in had a couple problems similar to the above. For the most part, I used a proof by induction…
Vincent
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$\sum^{\infty}_{n=1} \log(\frac{n+a+b}{n+a} \times \frac{n+b}{n+b+1})$

Prove $$\sum^{\infty}_{n=1} \log(\frac{n+a+1}{n+a} \times \frac{n+b}{n+b+1})=\log\frac{1+b}{1+a}$$ Hints/Answers are appreciated
the8thone
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