9 votes
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How to evaluate this sum $\sum_{n=1}^{\infty} \frac{(-1)^n}{(n^2 + 3n + 1)(n^2 - 3n + 1)}$

Let's use a standard way of evaluation of such sums. Denoting $S$ the desired sum, we consider the integral in the complex plane along a big circle $C_R$ with the radius $R$ $$I_C=\oint_C\frac\pi{\sin\...
Svyatoslav's user avatar
6 votes

How to evaluate this sum $\sum_{n=1}^{\infty} \frac{(-1)^n}{(n^2 + 3n + 1)(n^2 - 3n + 1)}$

A more elementary (high-school level) table-based proof that dispenses with complex analysis runs as follows. Starting from: $$x^2+3x+1 =\left(x-\frac{1}{2} \left(-\sqrt{5}-3\right)\right) \left(x-\...
user12030145's user avatar
  • 1,036
3 votes
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finding a tight scaling bound (in terms of the Big-O notation) of a function of an infinite sum of $1/n^2$.

The general method is to use the Euler-Maclaurin formula. In your case, you can notice that: $$ \frac1{n^2}\sim\frac1{n}-\frac1{n+1} $$ so: $$ \sum_{n=N}^\infty\frac1{n^2}\sim \sum_{n=N}^\infty\frac1{...
LPZ's user avatar
  • 2,580
3 votes
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For $a>1$ and a fixed $N$ does $\sum\limits_{n=1}^\infty \prod\limits_{k=0}^{n-1} \left(1- \frac{a}{a+N +k} \right)$ converge?

Multiply your series by $$\Gamma(N)\Gamma(a)/ \Gamma(a+N)$$ and your series becomes $$\sum_{n=1}^{\infty}\frac{\Gamma(a)\Gamma(N)\prod_{k=0}^{n-1} (N+k)}{\Gamma(a+N)\prod_{k=0}^{n-1} (a+N+k)}=\sum_{n=...
Kroki's user avatar
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3 votes

Any simplification of $\sum_{k=0}^{n}\binom{n}{k} \frac{(-1)^{k}}{(N-k)^2}$

In order to evaluate where $N\gt n$ $$\sum_{k=0}^n \frac{(-1)^k}{(N-k)^2} {n\choose k}$$ we introduce the function $$f(z) = \frac{n! (-1)^n}{(N-z)^2} \prod_{q=0}^n \frac{1}{z-q}.$$ This has the ...
Marko Riedel's user avatar
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2 votes

Why $n! \sum_{k=0}^n \frac{1}{(n-k)!} = n! \sum_{k=0}^n \frac{1}{k!}$?

Let's see what happens for Sigma $ n!\sum_{k=0}^n \frac{1}{(n-k)!} = n! \sum_{k=0}^n \frac{1}{k!} $ its suffice to show $$\sum_{k=0}^n \frac{1}{(n-k)!} = \sum_{k=0}^n \frac{1}{k!}$$ so put the ...
Khosrotash's user avatar
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2 votes
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What is the 1-case closed form for $\sum_{i = 1}^{x} \lfloor \frac{i - r}{d}\rfloor$?

By modifying range of $i$ at the two ends: supplementing $\left\lfloor\frac{1-r}{d} \right\rfloor d \le i-r < 1-r$, where $\left\lfloor\frac{i-r}{d} \right\rfloor = \left\lfloor\frac{1-r}{d} \...
peterwhy's user avatar
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2 votes
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Convergence of the series $f(a_n)a_n$

Suppose $\sum a_n $ absolutely converges. So $a_n\rightarrow0$ as $n\rightarrow \infty$. Then , by definition : $$\forall \epsilon>0, \exists N\in\mathbb{N} :n>N\implies|a_n|<\epsilon$$ So ...
J.Dmaths's user avatar
  • 662
2 votes
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Differences between sums of reciprocals of primes and products thereof.

Since $$\sum_{p < q \leq N}{\frac{1}{pq}} = \frac{1}{2}\left[\left(\sum_{p \leq N}{\frac{1}{p}}\right)^2-\sum_{p \leq N}{\frac{1}{p^2}}\right],$$ and since $\sum_p{\frac{1}{p}}=\infty$ and $\sum_{p}...
Aphelli's user avatar
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1 vote
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approximate a sum

I wouldn't expect there to be a simple exact answer unless some lucky coincidence happens. But it's not so hard to approximate. First of all, if we replace $\frac{x+1}{(x-1)x}$ with $\frac1x$ then the ...
Gareth McCaughan's user avatar
1 vote

Sums and Products over sets of sets

I assume the product in the sum was meant to be $\prod_{j\in A_i}x_j$. As long as some indexing set $I$ (possibly a set of sets) and the corresponding value for each index is something compatible with ...
Anush Rao's user avatar
1 vote
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Attempt at creating a formula relating debt, payments and interest

I think you will find it easier to work backwards so with $D_n$ being the amount outstanding with $n$ months remaining, starting with $D_0=0$, and you have $D_{n-1}=D_{n}(1+r)-p$ since you make a ...
Henry's user avatar
  • 156k
1 vote

Attempt at creating a formula relating debt, payments and interest

Let’s assume the payments are made at the end of each monthly period. At the end of month $1$ you owe a total of $D(1+r)-p$ At the end of month $2$ you owe a total of $\left(D(1+r)-p\right)(1+r)-p$ At ...
David Quinn's user avatar
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1 vote
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Proving Series has no limit

You can use some inequality to show $\sum_{n=1}^{\infty}n\sqrt[6]{n}(\sqrt[3]{n+1}-\sqrt[3]{n})$ diverge. $$n\sqrt[6]{n}(\sqrt[3]{n+1}-\sqrt[3]{n})=\\n\frac{\sqrt[6]{n}}{(\sqrt[3]{(n+1)^2}+\sqrt[3]{n(...
Khosrotash's user avatar
  • 24.6k
1 vote
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How do we evaluate the hypergeometric type series where $\delta_{n}=\large \int_{_{0}}^{^{N}} {{u} \choose {n}}{{N-u} \choose {N-n}} \textit{ du }$

On the surface, since $\sum_{n=0}^N{{u} \choose {n}}{{N-u} \choose {N-n}}=1$ (there is only one nonezero element with a value of $1$), the answer is $B$, i.e. $\sum_{n=0}^N\delta_n=N$. See this post ...
Math-fun's user avatar
  • 9,497
1 vote

how to calculate $\sum\limits_{k=1}^{+\infty }{\arctan \frac{1}{1+k^{2}}}$

Note that by the Weirestrass factorsation of $\sin(z)$, $$\sinh(\pi z) = \pi z\prod_{k=0}^{\infty} \left(1+\frac{z^2}{(k+1)^2}\right)$$ Thus the expression you left with, becomes $$\frac{i}2 \ln\left(\...
Sahaj's user avatar
  • 2,313
1 vote

how to calculate $\sum\limits_{k=1}^{+\infty }{\arctan \frac{1}{1+k^{2}}}$

It's good you did all that ground work. Let's rewrite $$\frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=0}^{n}{\left( \frac{\left( k+1 \right)^{2}+1-i}{\left( k+1 \...
user43208's user avatar
  • 7,999

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