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0 votes
2 answers
200 views

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

Question: how to calculate $$\sum\limits_{k=1}^{+\infty }{\arctan \frac{1}{1+k^{2}}}$$ My attempt Let $\arctan \theta =\frac{i}{2}\ln \left( \frac{i+\theta }{i-\theta } \right)$ $$S=\sum\limits_{k=1}^{...
Mods And Staff Are Not Fair's user avatar
1 vote
0 answers
102 views

Relative prime density of $f(n)$

Relative prime density of $f(n)$ Definition Number of primes of $f(n)$ Using the prime counting function $\pi(n)$, we can define the number of primes $\pi_{nbr}$ of a function $f(n)$ from $n=1$ to $...
François Huppé's user avatar
1 vote
1 answer
2k views

Changing upper limit of summation

I have a formula where both summation and product are involved \begin{align*} f(n-1)=\sum_{m=0}^{n-1}\sum_{j=0}^{m}\frac{\binom{m}{j}(-1)^j\mathrm{i}^m (a+k)^{2m-n+1}}{m!(2 a)^m} \prod_{p=1}^{n-1}(2m-...
shabbir's user avatar
  • 43
2 votes
1 answer
59 views

Is $\lim\limits_{n\to\infty} \sum\limits_{i=1}^n \prod\limits_{j=n-i+1}^n a_j=0$?

Let $\{a_n\}_{n\geqslant1}$ be a sequence of numbers in $(0,1)$ such that $a_n\to 0$ but $\sum\limits_{n\geqslant 1}a_n=\infty$ . Is it true that $$\lim\limits_{n\to\infty}\ \sum\limits_{i=1}^n\ \...
Qwerty's user avatar
  • 6,195
1 vote
1 answer
45 views

Limit of products in $x_n = 1-An^{-\alpha}$ and their summation

Suppose that we have $A >0, \alpha >0$, and for each $n$, define $x_n = 1-An^{-\alpha}$ such that for large $n$ we have $x_n \in (0,1)$. Also, define the product sequence, $y_n = \prod_{i=0}^n ...
Coleman's user avatar
  • 11
1 vote
3 answers
121 views

Finding limit of a product.

Prove:$$\lim_{n \to\infty }\frac{1}{n}\left[\prod_{i=1}^{n}(n+i) \right ]^{\frac{1}{n}}=\frac{4}{e}$$ I tried using Squeeze Theorem but can't go beyond $1<L<2$. $$\lim_{n\to\infty} \left( 1 + ...
ZSMJ's user avatar
  • 1,206