How do you use the integral test to determine the convergence or divergence of $1+1/sqrt2+1/sqrt3+1/sqrt4+...$ ?

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Answer 1 · 2017-05-08T17:43:14

The series is divergent and therefore has no finite sum.

This is defined by the formula

$t_n = 1/sqrt(n)$

Therefore, if the integral

$int_1^(oo) 1/sqrt(n) dn$

Has a calculable value, then the series converges. The integral can be rewritten as

$S = lim_(t->oo) int_1^t n^(-1/2) dn$

$S = [2n^(1/2)]_1^t$

$S = lim_(t->oo) 2sqrt(t) - lim_(t->oo) 2(1)^(1/2)$

The first limit obviously has value of $oo$ , therefore the series diverges and has no finite sum.

Hopefully this helps!

How do you use the integral test to determine the convergence or divergence of 1+1/sqrt2+1/sqrt3+1/sqrt4+...1+1/sqrt2+1/sqrt3+1/sqrt4+...?