MHB Series Convergence Or Divergence

[tmt1]

I have

$$\sum_{n = 2}^{\infty} \frac{(lnn)^ {12}}{n^{\frac{9}{8}}}$$

I'm trying the limit comparison test, so I let $$ b = \frac{1}{n^{\frac{9}{8}}}$$ and $a = \sum_{n = 2}^{\infty} \frac{(lnn)^ {12}}{n^{\frac{9}{8}}}$

$\frac{a}{b} = (lnn)^ {12}$ therefore I know the limit of this as n approaches infinity would be infinity.

Since $\infty > 0$, $a$ should converge if $b$ converges or diverge if $b$ diverges.

Since b converges ($9/8 > 1$, so $ \frac{1}{n^{\frac{9}{8}}}$ should converge), then a must converge.

Therefore, the original series converges. Is this correct?

 

[Prove It]

I have

$$\sum_{n = 2}^{\infty} \frac{(lnn)^ {12}}{n^{\frac{9}{8}}}$$

I'm trying the limit comparison test, so I let $$ b = \frac{1}{n^{\frac{9}{8}}}$$ and $a = \sum_{n = 2}^{\infty} \frac{(lnn)^ {12}}{n^{\frac{9}{8}}}$

$\frac{a}{b} = (lnn)^ {12}$ therefore I know the limit of this as n approaches infinity would be infinity.

Since $\infty > 0$, $a$ should converge if $b$ converges or diverge if $b$ diverges.

Since b converges ($9/8 > 1$, so $ \frac{1}{n^{\frac{9}{8}}}$ should converge), then a must converge.

Therefore, the original series converges. Is this correct?

First of all, you should note that any finite sum always gives a finite value.

For very large values of $\displaystyle \begin{align*} n \end{align*}$ (i.e. $\displaystyle \begin{align*} n > 2.661334067039 \cdot 10^{603} \end{align*}$) we have $\displaystyle \begin{align*} \left[ \ln{ \left( n \right) } \right] ^{12} < n^{\frac{1}{16}} \end{align*}$, so

$\displaystyle \begin{align*} \sum_{\textrm{Very large }n}^{\infty}{ \frac{\left[ \ln{ \left( n \right) } \right] ^{12}}{n^{\frac{9}{8}}} } &< \sum_{\textrm{Very large }n}^{\infty}{ \frac{n^{\frac{1}{16}}}{n^{\frac{9}{8}}} } \\ &= \sum_{\textrm{Very large }n}^{\infty}{ \frac{1}{n^{\frac{17}{16}}} } \end{align*}$

which is a convergent p series as $\displaystyle \begin{align*} \frac{17}{16} > 1 \end{align*}$.

Thus the first $\displaystyle \begin{align*} 2.661334067039\cdot 10^{603} \end{align*}$ terms will give a finite value, and we have shown that the remaining infinite sum converges by comparison, so that means $\displaystyle \begin{align*} \sum_{n = 2}^{\infty}{ \frac{\left[ \ln{\left( n \right) } \right] ^{12}}{n^{\frac{9}{8}}} } \end{align*}$ is convergent.