- #1
rocomath
- 1,755
- 1
[tex]\sum_{n=2}^{\infty}\frac{1}{(lnn)^{n}}[/tex]
If I treat it as a geometric series, then when n=3, r is smaller than 1
[tex]\sum_{n=2}^{\infty}(\frac{1}{ln3})^3[/tex]
What matters is that it's ultimately decreasing, would that be the correct approach? Even problem, no answer!
Thanks.
If I treat it as a geometric series, then when n=3, r is smaller than 1
[tex]\sum_{n=2}^{\infty}(\frac{1}{ln3})^3[/tex]
What matters is that it's ultimately decreasing, would that be the correct approach? Even problem, no answer!
Thanks.
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