12 votes

trying to solve the nested sum $\sum_{n=1}^{\infty} \frac{1}{1^2+2^2+\dots+n^2} $

We have \begin{align} \sum_{n=1}^\infty\left(\frac{1}{n}+\frac{1}{n+1}-\frac{4}{2n+1}\right) & = \sum_{n=1}^\infty\left(\frac{1}{n}-\frac{2}{2n+1}+\frac{1}{n+1}-\frac{2}{2n+1}\right)\\ &= \...
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  • 11.7k
5 votes
Accepted

Evaluate $\lim_{n\to\infty}\sum_{k = n^3}^{(n+1)^3}\frac{1}{\sqrt[3]{k^2 +4k}}$

Define $$S(n):= \sum\limits_{k = n^3 }^{(n + 1)^3 } {f(k)}$$ I shall follow Gary's hint; but provide more details. Consider $f: \Bbb N \to \Bbb R$ given by $$f(k) = \frac{1}{(k^2 + 4k)^{1/3}}$$ Note ...
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5 votes
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Is there a closed form of $x_2x_3\cdots x_n + x_1x_3\cdots x_n + \dots + x_1x_2\cdots x_{n-1}$?

There is this... $$ A_n = \big(x_1x_2\cdots x_n\big)\;\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right) $$ for the case all $x_i \ne 0$. If one of them is zero, say $x_1 = 0$, then $A_n = ...
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  • 97.4k
5 votes

Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$

We seek to calculate $$\begin{align}\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx&= \frac{1}{2}+\frac{1}{4}-\frac{1}{6}-\frac{1}{8}+\frac{1}{10}+\...
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3 votes

Evaluate a kronecker symbol sum: $\sum\limits_{n=1}^\infty \frac {\big(\frac n x\big)}n$ and $\sum\limits_{n=1}^\infty \frac {\big(\frac xn\big)}n$

These are values at $s = 1$ of the $L$ functions $L(\chi, s)$ attached to the quadratic characters $\chi(n) = \left(\frac m n\right)$. Depending on the parity of $\chi$ (i.e. whether $\chi(-1)$ is ...
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  • 21k
3 votes
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Express $\frac{1}{2n + 1}\,,\,n\in\mathbb{N}$ as a series of the form $\sum_{m = 0}^\infty\frac{a_m n!}{(n + m)!}\,.$

Let $(a_m)_{m\in\mathbb{N}}$ be a solution. Taking $n\to\infty$ (which can be done termwise), we get $a_0=0$. For $|x|<1$ the series $f(x)=\sum_{m=0}^\infty a_{m+1}x^m/m!$ converges, and we have $$\...
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3 votes

Can you calculate $1+2+3+4+\ldots+k$ geometrically?

Actually, think of it not as a triangle, but as a trapezium (or a trapezoid if you are in the US). The shape has a top length of $1$, not $0$, hence it is not a triangle. You probably understand it ...
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3 votes

trying to solve the nested sum $\sum_{n=1}^{\infty} \frac{1}{1^2+2^2+\dots+n^2} $

There are already two excellent answers showing how to evaluate the series correctly. I want to focus on how your attempt is incorrect. We'll also see another way to evaluate the series as a result, ...
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3 votes

Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$

Let $$a = \int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$$ Modified solution (now the main solution) We have two choices: integral first then sum or vice ...
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2 votes

The formula for $g\frac{d}{dg}g\frac{d}{dg}...g\frac{d}{dg}f(g)$

I do not prove the formula, but I relate your expression with known quantitites. A recursion yields $$\Big(x\frac{\mathrm{d}}{\mathrm{d}x}\Big)^nf(x) = \sum_{k=0}^n a_{n,k}x^kf^{(k)}(x),$$ where $a_{0,...
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1 vote
Accepted

Double summation - How to solve for $\sum_{j=i}^n 1?$

$$\begin{align} \sum_{i=1}^n \sum_{j=i}^n 1 &\stackrel{1.}= \sum_{i=1}^n (n+1-i) \\ &\stackrel{3.}= (n+1)\sum_{i=1}^n 1 - \sum_{i=1}^n i \\ &\stackrel{1., 2.}= (n+1)n - n(n+1)/2 \\ &\...
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1 vote

Double summation - How to solve for $\sum_{j=i}^n 1?$

\begin{eqnarray} \sum_{j=i}^{n}1&=&\underbrace{1+1+\cdots+1}_{n-i+1\text{ times}}\\ &=&n-i+1\\ \sum_{i=1}^{n}\sum_{j=i}^{n}1&=&\sum_{i=1}^{n}n-i+1\\ &=&n(n+1)-\sum_{i=1}...
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1 vote

trying to solve the nested sum $\sum_{n=1}^{\infty} \frac{1}{1^2+2^2+\dots+n^2} $

Sorry, too long for a comment We can use integration, but we have to be careful while changing the order of operations. Dealing with diverging terms, we can easily get an additional finite ...
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  • 6,284
1 vote

Can you calculate $1+2+3+4+\ldots+k$ geometrically?

Thanks all for pointing out the whole picture is not a triangle, I focused too much on the picture made of dots, neglecting the fact that the number '1' is a 'square', but not a dot, so combining them ...
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1 vote

trying to solve the nested sum $\sum_{n=1}^{\infty} \frac{1}{1^2+2^2+\dots+n^2} $

Actually you do not need integrals. $$ \sum_{k=1}^{2N+1}\frac{(-1)^{k+1}}{k}=\sum_{n=0}^{N}\frac{1}{2n+1}-\sum_{n=1}^{N}\frac{1}{2n} $$ converges to $\log(2)$ by the Maclaurin series of $\log(1+x)$ ...
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