The other day I was playing around with infinite sums and I found that for any fixed, positive integer $k$ the equality $$\sum_{n=1}^\infty\frac{1}{n(n+k)} = \frac{1}{k}\sum_{n=1}^k\frac{1}{n}$$ I got this result by using partial fraction decomposition and then cancelling out everything but a finite number of sums. This led me to seek a similar finite sum expression […]

In a problem in my book, the following equality is there: $$\sum_{n=0}^\infty\Big( \sum_{k_i\ge 0, \sum_{i=1}^\infty ik_i=n}\frac{x^n}{\prod_{i=1}^\infty k_i!(i!)^{k_i}}\Big)=\sum_{k_i\ge 0}\Big(\frac{x^{\sum_{i=1}^\infty ik_i}}{\prod_{i=1}^\infty k_i!(i!)^{k_i}}\Big)$$ where the summation runs over all sequences $k_1,k_2\cdots$ of non-negative integers containing finitely many non-zero entries. It is not clear to me how to this equality is obtained. I did work out something like $\sum_{j=0}^\infty\sum_{k=0}^jf(j,k)=\sum_{k=0}^\infty\sum_{j=k}^\infty […]

For $d\geq1$ let $I^d=[0,1)^d$ denote the $d$-dimensional half-open unit cube and consider a finite sequence $x_1,\ldots,x_N\in{I}^d$. For a subset $J\subset{I}$, let $A(J,N)$ denote the number of elements of this sequence inside $J$, i.e. $$ A(J,N)=\left|\{x_1,\ldots,x_N\}\cap{J}\right|, $$ and let $V(J)$ denote the volume of $J$. The discrepenacy of the sequence $x_1,\ldots,x_N$ is defined as $$ D_N=\sup_{J}{\left|A(J,N)-V(J)\cdot{N}\right|}, […]

This question already has an answer here: What is the value of lim$_{n\to \infty} a_n$ if $\lim_{n \to \infty}\left|a_n+ 3\left(\frac{n-2}{n}\right)^n \right|^{\frac{1}{n}}=\frac{3}{5}$? 3 answers

I want to show that the sequence $$a_n=\frac{1}{n}+(-1)^n$$ does not converge to a limit. I know that if a sequence $\left( a_n\right)_{n \in \mathbb{N}}$ converges to a limit L, then $\forall\epsilon >0\;\;\exists N\in\Bbb N \;\text{such that}\;\forall n\geq N, \;\;|a_n-L|\leq\epsilon$, but I am not sure how to use this to prove a sequence that does not […]

This question was posted/originated after a failure of a more generic attempt here: Let $\alpha_n$ be a sequence of positive Reals. It is known that $$\alpha_n \sim \log(n)$$ Let $\beta_n$ be another sequence of positive Reals such that, $$\sum_{k = 1}^n\beta_k \sim \log(n)$$ Can we say/prove that $$\frac{\sum_\limits{k = 1}^n\alpha_k\beta_k }{\sum_\limits{k = 1}^n\beta_k} \sim \frac{1}{2}\log(n)$$ […]

I was testing out a few summation using my previous descriped methodes when i found an error in my reasoning. I’m really hoping someone could help me out. The function which i was evaluating was $\sum_{n=1}^{\infty} n\ln(n)$ which turns out to be $-\zeta'(-1)$. This made me hope i could confirm my previous summation methode for […]

Plotting the series $$\displaystyle y = \sum_{k} \frac{\sin kx }{k}$$ In the limit it would look like Taking a finite number of terms, I want to understand what is the reason for the jiggling at the extremes, while there the jiggling in the middle is so small its not noticable. I truncated the sum to […]

Evaluate $$\sum_{n=1}^\infty \frac{n}{2^n}$$ My Work: $$\sum_{n=1}^\infty \frac{n}{2^n} = \sum_{n=1}^\infty n \left(\frac{1}{2}\right)^n$$ If we denote $f(x) = \sum_{n=1}^\infty nx^n$ then we wish to evaluate $f(1/2)$. Now, $$\sum_{n=1}^\infty nx^n = x \sum_{n=1}^\infty nx^{n-1} = x\sum_{n=1}^\infty (x^n)’ = x\left(\sum_{n=1}^\infty x^n\right)’ = x \left(\frac{-x}{1-x}\right)’ = \frac{-x}{(1-x)^2}$$ Applying $x=1/2$ we get the wrong result of $-2$. Where is my mistake?

Sorry for this uninteresting question but hopefully someone can provide some help. Is there a way to simplify the following expression? $$\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{v}\left(\frac{n-m-v}{n}\right)^{r}\displaystyle \frac{m}{m+v}$$ $$-\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{v}\left(\frac{n-m-v}{n}\right)^{r}$$ This is a hint for a problem, but I don’t know how to proceed. UPDATE: Would it bother you if I add the original problem?. Maybe some context is needed in […]

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