How do you use the integral test to determine if $\infty \sum n = 2 \frac{1}{n \sqrt{ln n}}$ $sum_(n=2)^oo 1/(nsqrtlnn)$ from $[2, \infty)$ $[2,oo)$ is convergent or divergent?

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Answer 1 · 2016-11-21T16:01:47

$n \sum k = 2 \frac{1}{k {log}_{e} (k)}$ $sum_(k=2)^n1/(k log_e(k))$ is divergent

Calling $f (x) = \frac{1}{x {log}_{e} (x)}$ $f(x)=1/(x log_e(x))$ we have that

$\int_{2}^{n} f (x) d x \leq n \sum k = 2 \frac{1}{k {log}_{e} (k)}$ $int_2^n f(x)dx le sum_(k=2)^n1/(k log_e(k))$ and

$\int_{2}^{n} f (x) d x = {log}_{e} (\frac{{log}_{e} n}{{log}_{e} 2})$ $int_2^n f(x)dx=log_e((log_e n)/(log_e 2))$ but

$lim_(n->oo)log_e(n)=oo$ so

$lim_(n->oo)sum_(k=2)^n1/(k log_e(k))=oo$

How do you use the integral test to determine if sum_(n=2)^oo 1/(nsqrtlnn)∞∑n=21n√lnnsum_(n=2)^oo 1/(nsqrtlnn) from [2,oo)[2,∞)[2,oo) is convergent or divergent?