Root Test/Comparison for $\sum_{k=1}^{\infty}\frac{(-3)^{k+1}}{4^{2k}}$

[ineedhelpnow]

comparison or root test? for testing convergence/divergence $\sum_{k=1}^{\infty}\frac{(-3)^{k+1}}{4^{2k}}$

 

[Euge]

comparison or root test? for testing convergence/divergence $\sum_{k=1}^{\infty}\frac{(-3)^{k+1}}{4^{2k}}$

You can use the ratio test. Rewriting the summands as $-3 (-3/16)^k$ for each $k$, you find that $\lim_{k \to \infty} |a_{k+1}/a_k| = 3/16 < 1$, where $a_k = -3 (-3/16)^k$. So by the ratio test, your series converges. In general, a geometric series $\sum_{k = 1}^\infty ar^{k-1}$ converges if $|r| < 1$ and diverges if $|r| > 1$.

 

[Nono713]

The root test works okay too...