Does the Comparison Test Determine Convergence or Divergence of Series?

[karush]

Use the comparison test to determine if the series series convergences or divergences
$$S_{6}=\sum_{n=1}^{\infty} \dfrac{1}{n^2 \ln{n} -10}$$
ok if i follow the example given the next step alegedly would be...
$$\dfrac{1}{n^2 \ln{n} -10}<\dfrac{1}{n^2 \ln{n}}$$
$\tiny{242 UHM}$

 

[skeeter]

note ...

$n^2\ln(n)-10 < n^2\ln(n) \implies \dfrac{1}{n^2\ln(n) - 10} > \dfrac{1}{n^2\ln(n)}$

$\displaystyle \sum \dfrac{1}{n^2\ln(n)} < \sum \dfrac{1}{n^2}$ converges ... so the comparison above isn't going to fly.

 

[karush]

oh ..

why put the $\sum $ back in ??

 

[skeeter]

oh ..

why put the $\sum $ back in ??

The goal is to compare series, not just the nth term.Let's compare \(\displaystyle \sum \frac{1}{n^2\ln(n)-10}\) with the known convergent series \(\displaystyle \sum \frac{1}{n^2}\)

if \(\displaystyle \sum \frac{1}{n^2\ln(n)-10} < \sum \frac{1}{n^2}\) for large enough $n$, then we can say \(\displaystyle \sum \frac{1}{n^2\ln(n)-10}\) converges

first off, we need to find what values of $n$ make \(\displaystyle \frac{1}{n^2\ln(n)-10} < \frac{1}{n^2}\)

for that inequality to be true ...

$n^2\ln(n) - 10 > n^2$

$n^2\ln(n) - n^2 > 10$

$n^2[\ln(n)-1] > 10 \implies n \ge 5$

so, \(\displaystyle \sum_{n=5}^\infty \frac{1}{n^2\ln(n)-10} < \sum_{n=5}^\infty \frac{1}{n^2} \implies \sum_{n=5}^\infty \frac{1}{n^2\ln(n)-10}\) converges.