MHB Does the Series $\sum_{n=2}^{\infty} (-1)^n \frac{4}{5\ln{n}}$ Converge?

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The series $\sum_{n=2}^{\infty} (-1)^n \frac{4}{5\ln{n}}$ converges by the alternating series test. The terms $\frac{4}{5\ln{n}}$ decrease as $n$ increases, satisfying the conditions for convergence. Since $\ln(n)$ is a monotonically increasing function, its reciprocal is monotonically decreasing. Thus, the series meets the criteria for convergence established by the alternating series test. Overall, the series converges due to the behavior of its terms.
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$\tiny{10.6.10}\\ $
$\textsf{ converge or diverge?}\\$
\begin{align*}\displaystyle
S_n&= \sum_{n=2}^{\infty} (-1)^n \frac{4}{5\ln{n}}\\
&\frac{4}{5} \sum_{n=2}^{\infty} (-1)^n
\frac{1}{\ln{n}}=
\end{align*}
?
??
?

$\textit{converges: alternating series test}$
 
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karush said:
$\tiny{10.6.10}\\ $
$\textsf{ converge or diverge?}\\$
\begin{align*}\displaystyle
S_n&= \sum_{n=2}^{\infty} (-1)^n \frac{4}{5\ln{n}}\\
&\frac{4}{5} \sum_{n=2}^{\infty} (-1)^n
\frac{1}{\ln{n}}=
\end{align*}
?
??
?

$\textit{converges: alternating series test}$

Yes it converges as ln(n) is a decreasing function.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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