Fix a positive number α and consider the series∑k=1∞1(k+1)[ln⁡(k+1)]α

Consider the function 1(x+1)[ln⁡(x+1)]α for α&gt;0. Then 1(k+1)[ln⁡(k+1)]α is positive, decreasing, continuous on [2,∞). Therefore, by integral test, we have∑k=1∞1(k+1)[ln⁡(k+1)]αconverges if and only if∫1∞1(x+1)[ln⁡(x+1)]αdxconverges. On the other hand, if α&gt;1, then∫1∞1(x+1)[ln⁡(x+1)]αdx=11−αln⁡(x+1)1−α∣∞1&lt;∞and it diverges if 0&lt;α≤1. Therefore, the series converges when α&gt;1 and diverges when 0&lt;α&lt;1Also, for α≤0, we have1(k+1)[ln⁡(k+1)]α≥1k+1and the harmonic series ∑k=1∞1k+1 diverges. By comparison test, ∑k=1∞1(k+1)[ln⁡(k+1)]α diverges when α≤0

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gabolzm6d

Answered question

2022-04-06

Fix a positive number

α

and consider the series

\sum_{k = 1}^{\infty} \frac{1}{(k + 1) {[\ln (k + 1)]}^{α}}

Answer & Explanation

libertydragonrbha

Beginner2022-04-07Added 15 answers

Consider the function

\frac{1}{(x + 1) {[\ln (x + 1)]}^{α}}

for

α > 0

. Then

\frac{1}{(k + 1) {[\ln (k + 1)]}^{α}}

is positive, decreasing, continuous on

[2, \infty)

. Therefore, by integral test, we have

\sum_{k = 1}^{\infty} \frac{1}{(k + 1) {[\ln (k + 1)]}^{α}}

converges if and only if

\int_{1}^{\infty} \frac{1}{(x + 1) {[\ln (x + 1)]}^{α}} d x

converges. On the other hand, if

α > 1

, then

\int_{1}^{\infty} \frac{1}{(x + 1) {[\ln (x + 1)]}^{α}} d x = \frac{1}{1 - α} {\ln (x + 1)}^{1 - α} ∣^{\infty}_{1} < \infty

and it diverges if

0 < α \leq 1

. Therefore, the series converges when

α > 1

and diverges when

0 < α < 1

Also, for

α \leq 0

, we have

\frac{1}{(k + 1) {[\ln (k + 1)]}^{α}} \geq \frac{1}{k + 1}

and the harmonic series

\sum_{k = 1}^{\infty} \frac{1}{k + 1}

diverges. By comparison test,

\sum_{k = 1}^{\infty} \frac{1}{(k + 1) {[\ln (k + 1)]}^{α}}

diverges when

α \leq 0

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