# Does the series Sum (k=2, oo) 1/((ln(k))^ln(k)) converge diverge or?

I'd like some help to decide if the series Sum (k=2, oo) 1/((ln(k))^ln(k)) converge or diverge.

Thank you

### 3 Answers

- hustolemynameLv 61 decade agoFavorite Answer
a[k] = 1/ln(k)^ln(k)

a[k+1] = 1/ln(k(1+1/k))^ln(k+1)

= 1/( ln(k) + ln(1+1/k))^ln(k+1)

= 1/ ( ln(k)^ln(k+1) ( 1+ (ln(1+1/k)/ln(k) )^ln(k+1)

a[k]/a[k+1] =ln(k)(1+(ln(1+1/k)/ln(k) ) ^ ln(k+1) > ln(k)

after the first term (or after the first few terms if we had some other base) ln(k) > 1

in particular after 100 terms

a[k+1]/a[k] < 1/ln[k] < 1/ln(100) < 1/4

therefore the series converges

Edit -

In view of MathMark's answer posted about the same time I had a look at the sum in excel.

I think its interesting that after the first 95 terms there are

135 terms < 1/10^3 totalling < 0.135

300 terms < 1/10^4 totalling < 0.0300

640 terms < 1/10^5 totalling < 0.00640

1348 terms < 1/10^6 totalling < 0.001348

2770 terms < 1/10^7 totalling < 0.0002770

5580 terms < 1/10^8 totalling < 0.00005580

11126 terms< 1/10^9 totalling < 0.000011126

21863 terms <1/10^10totalling <0.0000021863

This doesn't prove anything of course, but its enough to convince me I don't have to spend more time looking for an error in my logic

For info it has all the appearance of converging to 5.716970544

- guerreroLv 43 years ago
The shrink of (-a million)^n (n+a million)/sqrt(n) as n-->inf DNE, so the endless sequence diverges to boot. many times, the endless sum of phrases can purely exist if the shrink of the phrases themselves is 0 (even with the certainty that the shrink of the sums of the phrases need not be 0 of direction).

- 1 decade ago
we can compare it to 1/(lnk)^(k)

ln(k) < k

so

ln(k) ^ ln(k) < (ln k) ^ k

so

1/ (ln k)^ln k > 1/ (ln k) ^ k

Since 1 / (ln k) ^ k diverges, so does this series.