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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Dirac and sign errors in odd number of places

I recently came across this Dirac anecdote. How does one prove that signs must have been wrong in odd number of places? Does this follow from parity? Now for example if I calculate 1+2+3+4 I get 10 and if I make sign mistake in even number of…
danny
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An equality: from one sum to 2 sums.

I have an equality: $$ \ddot{a}_{30} = \frac{1}{0.75}\left( \sum_{k=0}^{\infty} \left(\frac{1}{1.06}\right)^k \left( 1 - \frac{30+k}{120}\right) \right)=\left( \sum_{k=0}^{\infty} \left(\frac{1}{1.06}\right)^k - \frac{1}{90} \sum_{k=0}^{\infty}…
Philip
  • 313
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Prove that $\sum_{j=0}^n \sum_{k=0}^j p_k q_{j-k} r_{n-j} = \sum_{k=0}^n \sum_{j=k}^n p_k q_{j-k} r_{n-j}$

I encounter this problem when proving that $\Bbb R[[X]] := (\Bbb R^\Bbb N,+,\cdot)$ is actually a formal power series ring over $\Bbb R$. Let $(p_n \mid n \in \Bbb N), (q_n \mid n \in \Bbb N), (r_n \mid n \in \Bbb N)$ be sequences in $\Bbb R$.…
Akira
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Can you help me with this sum?

Why isn't there a proper solution on internet on this sum: $$\sum_{i=1}^{n} \frac{1}{i}$$ Can you help me with your solution?
Adnan C
  • 63
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Sum of n*constant^n from 1 to N

What is the value of the sum for n=1 to n=N of n * constant ^ n? I have considered Geometric but with the multiplication by n I cannot get this to work.
MrT
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Evaluating a finite summation

What is the general formula for the following sum: $$\sum^N_{n=1}\frac{n+1}{n}$$ where $N$ is a finite natural number.
M. Gilbert
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Property of summation

Very short question. Could you please explain me why $$\sum_{i=0}^{n-1} a = na$$ with $a$ a constant? I know that $$\sum_{i=1}^{n} a = na$$ but in my case the sum starts from zero and finishes for $(n-1)$. Thanks.
Kolmogorovwannabe
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How to read this math equation? $\sum_{i=0}^{n-2}(n-1-i)$

I stumbled upon this in a book and am uncertain how to read it. This is in the context of describing run-time complexity of a computer algorithm. $$\sum_{i=0}^{n-2}(n-1-i)$$
Taylor Clark
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Calculate sum $\sum_{n=1}^\infty {n\over3^{n-1}}$

Calculate sum $$\sum_{n=1}^\infty {n\over3^{n-1}}$$The result is $9\over 4$ but I don't know how to get that.
retne
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Summation of Near-Rising-Factorial

Is there an easy way to find the closed form of $$\scriptsize\sum_{r=1}^n (r+\tfrac {2010}{10})(r+2011)(r+2012)(r+2013)(r+2014)(r+2015)(r+2016)(r+2017)(r+2018)$$ without first knowing the answer? The answer is $$\scriptsize{\frac…
Hypergeometricx
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Evaluate sum $\sum_{k=0}^N \binom{N}{k} \prod_{j=1}^k [ (j-1)\lambda + (N-1)\mu ] \prod_{i=1}^{N-k} [(i-1)\lambda + (N-1)\gamma ]$

By calculating the first few terms of N, I think the sum $$\sum_{k=0}^N \binom{N}{k} \prod_{j=1}^k [ (j-1)\lambda + (N-1)\mu ] \prod_{i=1}^{N-k} [(i-1)\lambda + (N-1)\gamma ]$$ is equal to $$ \prod_j^N [(j-1) \lambda + (N-1)\mu + (N-1)…
Calum Gabbutt
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Why is the accepted summation multiplication formula what it is?

$\sum_{n=0}^3 n = 6$ and $6^2 = 36$ I know the formula is $$\left( \sum_{n=0}^\infty a_n \right)\left( \sum_{n=0}^\infty b_n \right) = \sum_{n=0}^\infty \sum_{k=0}^n (a_n)(b_{n-k})$$ So: $$\left( \sum_{n=0}^\infty a_n \right)^2 = \sum_{n=0}^\infty…
Ethan Bartiromo
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How can I write this as a double sum?

I'm confused about this notation, can someone please explain how I can write $2 \sum_{1\leq j < k \leq N} P(F_j F_k)$ as a double sum. Would it be $\sum_{j=1}^k \sum_{k=n}^N P(F_j F_k)$? Also I am just confused about what a double sum would…
user477465
  • 720
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For arbitrary positive integral $n,$ find the sum $2^{n}+2^{n-1}+2^{n-2}+\ldots+2^{-n+1}+2^{-n}$

For arbitrary positive integral $n,$ find the sum $$2^{n}+2^{n-1}+2^{n-2}+\ldots+2^{-n+1}+2^{-n}.$$ My solution: $$\begin{align} 2^{n}+2^{n-1}+2^{n-2}+\ldots+2^{-n+1}+2^{-n}&=2^{n}[2^{0}+2^{-1}+2^{-2}+2^{-3}+\dots…
Tariro Manyika
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Why does the sum of probabilities of intersections of events equal this?

Why is it that $$ \sum_{k=1}^N P(A_k) = \sum_{j=k}P(A_j A_k)$$ I know the law of total probability but I'm not sure if this applies to that. I'm not told that the sequence of events are disjoint or independent so I'm just wondering how I would get…
user130306
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