$$\sum _{ n=10 }^{ \infty }{ (-1)^ n } \frac { n^ 2 }{ \ln(n) }$$
My attempt
Since this is an alternating series,
$$\lim _{ n\rightarrow \infty }{ \frac { n^ 2 }{ \ln(n) } } $$
diverges. So, the series diverges. Can anyone please verify this?
$$\sum _{ n=10 }^{ \infty }{ (-1)^ n } \frac { n^ 2 }{ \ln(n) }$$
My attempt
Since this is an alternating series,
$$\lim _{ n\rightarrow \infty }{ \frac { n^ 2 }{ \ln(n) } } $$
diverges. So, the series diverges. Can anyone please verify this?
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I don't have enough rep to comment, so I'll write this as an answer. Yes, you're correct.
Since $\lim_n \frac{n^2}{\log n} = \infty$, it follows that $\lim_n (-1)^n\frac{n^2}{\log n} \neq 0$. Thus, the series diverges.