Question

Determine whether the series converges or diverges:

$\sum_{n=3}^{\infty} \frac{7}{\sqrt{n+1} \ln(\sqrt{n+1})}$

Answer 1

Diverges

Answer 2

To determine whether the series $\sum_{n=3}^{\infty} \frac{7}{\sqrt{n+1} \ln(\sqrt{n+1})}$ converges or diverges, we can use the Limit Comparison Test.

Let $a_n = \frac{7}{\sqrt{n+1} \ln(\sqrt{n+1})}$. We can rewrite $\ln(\sqrt{n+1})$ as $\frac{1}{2} \ln(n+1)$, so $a_n = \frac{14}{\sqrt{n+1} \ln(n+1)}$.

For large $n$, $\sqrt{n+1} \approx \sqrt{n}$ and $\ln(n+1) \approx \ln n$. Thus, $a_n \approx \frac{14}{\sqrt{n} \ln n}$.

Consider the series $\sum_{n=2}^{\infty} \frac{1}{n^p (\ln n)^q}$. This series diverges if $p \le 1$ and $q \le 1$. In our case, we have $a_n \approx \frac{14}{n^{1/2} (\ln n)^1}$, where $p = 1/2 < 1$ and $q = 1$.

Let's use the Limit Comparison Test with $b_n = \frac{1}{\sqrt{n} \ln n}$. We know that $\sum_{n=2}^{\infty} \frac{1}{\sqrt{n} \ln n}$ diverges because $p = 1/2 < 1$.

Now, let's compute the limit:

$L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{14}{\sqrt{n+1} \ln(n+1)}}{\frac{1}{\sqrt{n} \ln n}} = \lim_{n \to \infty} \frac{14 \sqrt{n} \ln n}{\sqrt{n+1} \ln(n+1)} = \lim_{n \to \infty} 14 \sqrt{\frac{n}{n+1}} \cdot \frac{\ln n}{\ln(n+1)}$

We know $\lim_{n \to \infty} \sqrt{\frac{n}{n+1}} = \lim_{n \to \infty} \sqrt{\frac{1}{1+1/n}} = \sqrt{1} = 1$.

We also know $\lim_{n \to \infty} \frac{\ln n}{\ln(n+1)} = \lim_{n \to \infty} \frac{\ln n}{\ln(n(1+1/n))} = \lim_{n \to \infty} \frac{\ln n}{\ln n + \ln(1+1/n)} = \lim_{n \to \infty} \frac{1}{1 + \frac{\ln(1+1/n)}{\ln n}} = \frac{1}{1+0} = 1$.

So, $L = 14 \cdot 1 \cdot 1 = 14$.

Since $L = 14$ (a finite, positive number) and $\sum b_n = \sum \frac{1}{\sqrt{n} \ln n}$ diverges, by the Limit Comparison Test, the series $\sum a_n$ also diverges.

Therefore, the series diverges.