How many terms of the series [tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} [/tex] would you need to add to find its sum to within 0.01?(adsbygoogle = window.adsbygoogle || []).push({});

Here's what i got:

let [tex] f(n) = \frac{1}{n(ln\;n)^{2}} [/tex]. Since [tex] f(n) [/tex] is continuous, positive and decreasing for all n over the interval [tex] [2,\infty] [/tex], we can use the integral test to evaluate the series.

[tex]\sum^{\infty}_{n=2} \frac{1}{n(ln\;n)^{2}} = \lim_{b \to \infty} \int_{2}^{b} \frac{1}{n(ln\;n)^{2}} = \frac{1}{ln\;n} [/tex]

thus,

[tex] R_{n} \leq \frac{1}{ln\;n} [/tex]

since we want [tex] R_{n} \leq 0.01 [/tex],

[tex] \frac{1}{ln\;n} \leq 0.01 [/tex]

implying [tex] n = e^{100}[/tex].

but that can't be right...e^100 is way too big, isn't it? thanks in advance.

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# The integral test

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