$$\sum_{n = 2}^\infty \, \frac{n^3}{n^4 - 1}$$

This series looks like the harmonic series, and the -1 in the denominator makes it look like it's greater, term-by-term, than the terms of that series. Let's compare them:

$$\frac{n^3}{n^4 - 1} \stackrel{?}{\gt} \frac{1}{n}$$

$$\longrightarrow \; n^4 \gt n^4 - 1$$

This is true for all n, so the series diverges; all of its terms are greater than the terms of the divergent harmonic series.

$$\sum_{n = 1}^\infty \, \frac{e^{1/n}}{n}$$

This series looks like the harmonic series, and the e^1/n term in the numerator makes it look like it's greater, term-by-term, than that series. Let's compare

$$\frac{e^{1/n}}{n} \stackrel{?}{\gt} \frac{1}{n}$$

$$ \begin{align} \longrightarrow \; ne^{1/n} &\gt n \\ \\ e^{1/n} &\gt 1 \\ \\ e &\gt 1^n \end{align}$$

This is true for all n, so the series diverges; all of its terms are greater than the terms of the divergent harmonic series.

$$\sum_{n = 0}^\infty \, \frac{n + 5^n}{n + 6^n}$$

When n is large, the terms of the form "n" divide to 1. What's left is basically $\left( \frac{5}{6} \right)^n,$ which is a convergent geometric series. So let's try to show that, term-by-term, the terms of this series are less than those of that geometric series:

$$\frac{n + 5^n}{n+ 6^n} \stackrel{?}{\lt} \left( \frac{5}{6}\right)^n$$

$$ \begin{align} \longrightarrow \; n 6^n + 5^n6^n &\lt n 5^n + 6^{2n} \\ \\ n(6^n - 5^n) &\lt 6n(6^n - 5^n) \\ \\ n &\lt 6n \end{align}$$

This is true for all n > 0, so the series converges

$$\sum_{n = 0}^\infty \, \frac{2 + sin(n)}{10^n}$$

The sin(n) term in the numerator is always within -1 < sin(n) < 1, so the numerator fluctuates between 1 and 3. If we view that numerator as a small constant, then this is nearly a geometric series with $r = \frac{1}{10}.$ Let's compare it to the convergent geometric series with $r = \frac{1}{2}.$

$$\frac{2 + sin(n)}{10^n} \stackrel{?}{\lt} \left( \frac{1}{2}\right)^n$$

$$ \begin{align} \longrightarrow \; 2 \cdot 2^n + 2^n sin(n) &\lt 10^n \\ \\ 2^n (2 + sin(n)) &\lt 10^n \\ \\ 2 + sin(n) &\lt 5^n \\ \\ \end{align}$$

$$1 \lt 2 + sin(n) \lt 3$$

The inequality is true for all n ≥ 0, so the series converges.

$$\sum_{n = 1}^\infty \, \frac{4^{n + 1}}{3^n - 2}$$

For large values of n, the 2 has little effect, so this looks vaguely like the divergent harmonic series with $r = \frac{4}{3},$ and because the exponent in the numerator is one greater than that of the denominator, we further expect divergence. Compare this series to the divergent series with terms $\left(\frac{4}{3}\right)^n.$

$$\frac{4^{n + 1}}{3^n - 2} \stackrel{?}{\gt} \frac{4^n}{3^n}$$

$$ \begin{align} \longrightarrow \; 4^{n + 1}3^n &\gt 4^n(3^n - 2) \\ \\ \frac{4^{n + 1}}{4^n} &\gt \frac{3^n - 2}{3^n} \\ \\ 4 &\gt 1 - \frac{2}{3^n} \end{align}$$

This is true for all n ≥ 0, so the series diverges; all of its terms are greater than the terms of the divergent harmonic series.

$$\sum_{n = 1}^\infty \, \frac{n - 1}{n^2 \sqrt{n}}$$

If we disregard the 1 in the numerator, which has little effect for large n, the series has terms like (1/n)^3/2, which is a convergent p-series, so compare this series to the p-seris with terms (1/n)^3/2.

$$\frac{n - 1}{n^2 \sqrt{n}} \stackrel{?}{\lt} \left( \frac{1}{n} \right)^{3/2}$$

$$ \begin{align} \longrightarrow \; n\cdot n^{3/2} - n^{3/2} &\lt n^{5/2} \\ \\ n - 1 &\lt n \end{align}$$

This is true for all n ≥ 1, so the series converges; all of its terms are less than the terms of the convergent harmonic series.