V. Kaushik

activity

20172020

most citedEvaluation of Harmonic Sums with Integrals

7 citations · 8 across the 4 of their papers we have counts for

collaborators

4 papers

math.PR2020★ 1 cited

A Simple Multiple Integral Solution to the Broken Stick Problem

Vivek Kaushik

Regard the closed interval $[0, 1]$ as a stick. Partition $[0, 1]$ into $n + 1$ different intervals $I_{1}, \dots, I_{n + 1},$ where $n \geq 2,$ which represent smaller sticks. The cl…

math.CA2019

A Comment on the Sums $\sum_{n \in Z} \frac{( - 1 ) ^{nk}}{( an + 1 ) ^{k}}$

Vivek Kaushik

We recall a proof of Euler's identity $\sum_{n = 1}^{\infty} \frac{1}{n ^{2}} = \frac{π ^{2}}{6}$ involving the evaluation of a double integral. We extend the method to find Hurwitz Zeta ser…

math.CA2018

On Central Binomial Series Related to Zeta(4)

Vivek Kaushik

In this paper, we prove two related central binomial series identities: $B (4) = \sum_{n \geq 0} \frac{( n 2 n )}{2 ^{4 n} ( 2 n + 1 ) ^{3}} = \frac{7 π ^{3}}{216}$ and $C(4)=\sum_{n \in \mathbb{N}…

math.PR2017★ 7 cited

Evaluation of Harmonic Sums with Integrals

Vivek Kaushik, Daniele Ritelli

We consider the sums $S (k) = \sum_{n = 0}^{\infty} \frac{( - 1 ) ^{nk}}{( 2 n + 1 ) ^{k}}$ and $ζ (2 k) = \sum_{n = 1}^{\infty} \frac{1}{n ^{2 k}}$ with $k$ being a positive integer. We evaluate these sums…

A Simple Multiple Integral Solution to the Broken Stick Problem

A Comment on the Sums ∑n∈Z​(an+1)k(−1)nk​

On Central Binomial Series Related to Zeta(4)

Evaluation of Harmonic Sums with Integrals

A Comment on the Sums $\sum_{n \in Z} \frac{( - 1 ) ^{nk}}{( an + 1 ) ^{k}}$