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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Summation notation geometric sequence

I am perplexed as to how to solve this problem: $$\sum_{k=0}^{10} \left(\frac{-1}{4}\right)^k$$ The answer is $838861/1048576$ but I do not know how to get there.
Jinzu
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Showing that $\sum_{n\in N}\sum_{k=0}^nu_kT^n=\sum_{k\in N}\sum_{l\in N}u_kT^{k+l}$

We place ourselves in the ring of formal power series. Let $(u_k)$ be a family of natural integers. How do I prove rigorously the following equality ? $$\sum_{n\in N}\sum_{k=0}^nu_kT^n=\sum_{k\in N}\sum_{l\in N}u_kT^{k+l}$$ This is taken from a…
Minuteman
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Show that $\sum_{k=0}^{n} \sum_{r=k+1}^{n+1} \binom{n}{k} \binom{n+1}{r} = 2^{2n}$

I am trying to show: $$\sum_{k=0}^{n} \sum_{r=k+1}^{n+1} \binom{n}{k} \binom{n+1}{r} = 2^{2n}$$ Could I have a hint? I have tried rewriting $ \sum_{k=0}^{n} \sum_{r=k+1}^{n+1} \binom{n}{k} \binom{n+1}{r} = \sum_{k=0}^{n} \sum_{r=0}^{n-k}…
Oliver
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How to find $\sum_{n=0}^{\infty}e^{a n + b n^2}$?

I am trying to compute some quantity like a partition function for a harmonic oscillator type system and at the end of the computation I am coming across a sum that looks like, \begin{align} S=\sum_{n=0}^{\infty}e^{an +bn^2} \end{align} with $a,b…
Fermion
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Proving the identity $\frac{\sum_{i=1}^{l} (2i)^{2}}{2} = \sum_{i=1}^{l} 2(l-i)^{2}$.

In one of the lecture notes I have, there is an exercise in which I have to prove the following equality with sums: \begin{equation} \frac{\sum_{i=1}^{l} (2i)^{2}}{2} = \sum_{i=1}^{l} 2(l-i)^{2}. \end{equation} My approach:…
Math123
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Given some $n$, is it possible to select $k_i$ such that $3^n2^{k_0} + 3^n2^{k_1}+ 3^n2^{k_{2}} + \cdots + 3^n2^{k_i} = 2^m - 1 $

Given some $n$, is it possible to select $k_i$ such that the following is true? $$ 3^n2^{k_0} + 3^n2^{k_1}+ 3^n2^{k_{2}} + 3^n2^{k_3} + ... + 3^n2^{k_i} = 2^m - 1 $$ for $n=1$ it's trivial, for $n=2$ it's $$ 9 + 9 \times 2 + 9 * 4 = 64 - 1 $$ for…
alwaysmpe
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How to simplify this summation containing floor

Here's the summation: $$ S = \sum_{i=1}^{n}\left\lfloor\frac{n}{i}\right\rfloor $$ It can also be written as $$ S = \sum_{i=1}^{n}\left\lfloor\frac{n- (n \mod i)}{i}\right\rfloor $$ The answer will always be an integer, but I can't figure out how to…
Rushil Paul
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Double summation swap order when the inner index is function of the outer index

given this summation $\sum_{t=1}^T \sum_{s= \lfloor g(t) \rfloor}^t x_{t,s}$, where $g(t)$ is an increasing function satisfying $g(t)\leq t$ for all $t$, is it possible to swap the indexes of the two sums? I have been thinking it should be…
Alessio Russo
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How to show that $\sum_{n=M}^N a_n b_n= \sum_{k=M}^{N-1}s_k(b_k-b_{k+1})s_Nb_N-s_{M-1}b_M$

Suppose $a_n$ and $b_n$ are finite sequence of real numbers. Let $s_k = \sum_{n=1}^ka_n$ with onvention $s_0= 0$.Then show that $$\sum_{n=M}^N a_n b_n= \sum_{k=M}^{N-1}s_k(b_k-b_{k+1})+s_Nb_N-s_{M-1}b_M.$$ Trial: I know that $\sum_{n=M}^N a_n…
Argha
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Is it possible to divide summations across both sides of an equation?

For example, when $\sum$ goes from $i=0$ to $k$ on both sides: $$\sum_{i=0}^k n_i b^i \equiv \sum_{i=0}^k n_i \pmod{b-1}$$ Is it possible to divide out the ni on both sides? If not, are there any other ways to simplify this equation that I may be…
Matt Vukas
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Double Summation with an Exclusive Term

We can use $\sum_{i = 1}^{m}\sum_{j = 1}^{n} A_{ij}$ to denote the sum of all the elements in matrix $A_{m \times n}$. Now I would like to denote the sum of all the elements except $A_{rc}$. I come up with 2 options: $$ \sum_{\substack{i = 1 \\ i…
user1003454
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Summation Question

I am aiming to show the largest representable integer with n bits in base -2 encoding. However I do not really know how to work with summations. For example: 1001 in base -2 is -7: 1*(-2^3) + 0*(-2^2) + 0*(-2^1) + 1*(-2^0) which is -8 + 1 = -7. As…
Secret
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How do I evaluate the sum $\sum_{n=0}^{\infty}\frac{2^{n}}{2^{2^{n}}+1}$?

The series clearly converges, but I have no idea how to approach it using basic methods. How do I solve it?
FreshestCheese
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Is it true that if for two converging sums $\sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty b_n x^n$ then $\forall_{n≥0} a_n=b_n$?

Is it true that if $x\in\mathbb{R}$ for two converging sums $$\sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty b_n x^n,$$ then $$\forall {n≥0},a_n=b_n?$$ What if $x$ is not real but complex? Or any ring?
Damian Albrecht
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Distributivity property of summations when index is equal among summations

I Know that this equivalence is true, considering the distributivity property of summation: $$\sum \limits_{i=1}^{N}x_i g_{i,l}\sum \limits_{j=1}^{N}x_j g_{j,k}=\sum \limits_{i=1}^{N}\sum \limits_{j=1}^{N}x_{i}x_{j}g_{i,l}g_{j,k}$$ I was wondering…
CafféSospeso
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