Questions tagged [summation]

11,405 questions
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How close can $\sum_{k=1}^n \sqrt{k}$ be to an integer?

How close can $S(n) = \sum_{k=1}^n \sqrt{k}$ be to an integer? Is there some $f(n)$ such that, if $I(x)$ is the closest integer to $x$, then $|S(n)-I(S(n))|\ge f(n)$ (such as $1/n^2$, $e^{-n}$, ...). ...
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What is $\sum\limits_{i=1}^n \sqrt i\$?

What is $\sum\limits_{i=1}^n\sqrt i\$? Also I noticed that $\sum\limits_{i=1}^ni^k=P(n)$ where $k$ is a natural number and $P$ is a polynomial of degree $k+1$. Does that also hold for any real ...
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Trig sum: $\tan ^21^\circ+\tan ^22^\circ+…+\tan^2 89^\circ = ?$

As the title suggests, I'm trying to find the sum $$\tan^21^\circ+\tan^2 2^\circ+...+\tan^2 89^\circ$$ I'm looking for a solution that doesn't involve complex numbers, or any other advanced branch in ...
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How does one [easily] calculate $\sum\limits_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)}$?

How does one [easily] calculate $\sum\limits_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)}$, where $\mathrm{pop}(n)$ counts the number of bits '1' in the binary representation of $n$? Is there any trick ...
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Closed form for $\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}$

Please help me to find a closed form for the sum $$\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n},$$ where $H_n$ are harmonic numbers: $$H_n=\sum_{k=1}^n\frac{1}{k}=\frac{\Gamma'(n+1)}{n!}+\gamma.$$
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How to prove a formula for the sum of powers of $2$ by induction?

How do I prove this by induction? Prove that for every natural number n, $2^0 + 2^1 + ... + 2^n = 2^{n+1}-1$ Here is my attempt. Base Case: let $n = 0$ Then, $2^{0+1} - 1 = 1$ Which is true. ...
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Simplify $\left({\sum_{k=1}^{2499}\sqrt{10+{\sqrt{50+\sqrt{k}}}}}\right)\left({\sum_{k=1}^{2499}\sqrt{10-{\sqrt{50+\sqrt{k}}}}}\right)^{-1}$

Simplify $$\frac{\displaystyle\sum_{k=1}^{2499}\sqrt{10+{\sqrt{50+\sqrt{k}}}}}{\displaystyle\sum_{k=1}^{2499}\sqrt{10-{\sqrt{50+\sqrt{k}}}}}$$ I don't have any good idea. I need your help.
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Is this algebraic identity obvious? $\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1$

If $\lambda_1,\dots,\lambda_n$ are distinct positive real numbers, then $$\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1.$$ This identity follows from a probability calculation ...
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Mysterious identity

Playing around with Maple I found this identity $$\sum_{k=0}^{n-1}\frac{2k+1}{1-z^{2k+1}}=n\sum_{k=0}^{n-1}\frac{1}{1+z^{k}}$$ where $n$ is a positive integer, $z=\exp(\pi i/n)$. I was able to verify ...
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What are the applications of finite calculus

I'm reading through Concrete Mathematics [Graham, Knuth, Patashnik; 2nd edition], and in the section regarding Summation, they have a sub-section entitled "Finite and Infinite Calculus". In this ...
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Fibonacci-related sum

Related to this question Find a solution for $f\left(\frac{1}{x}\right)+f(x+1)=x$, what is this sum: $$\sum_{n=1}^{\infty}(-1)^n\left(\frac{F_n}{F_{n+1}}-\frac1{\phi}\right)$$ where $F_n$ is the $n$th ...
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Binomial Coefficients involving Prime Powers Minus 1

I would like to show the following is true; Let $p\in\mathbb{P}, \alpha,n\in\mathbb{N}$. Then $$p^\alpha\mid\sum_{k=1}^{n-1}\binom{(p^\alpha-1)n}{(p^\alpha -1)k}.$$ I've never worked with prime ...
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How find this $\sum_{n=1}^{\infty}\frac{H^3_{n}}{n+1}(-1)^{n+1}$

How Find this sum $$I=\sum_{n=1}^{\infty}\dfrac{H^3_{n}}{n+1}(-1)^{n+1}$$ where $H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$ My idea: since $$\dfrac{1}{n+1}(-1)^{n+1}=-\int_{-1}^{0}x^ndx$$...