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

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comparison or root test? for testing convergence/divergence $\sum_{k=1}^{\infty}\frac{(-3)^{k+1}}{4^{2k}}$
 
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ineedhelpnow said:
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 $|r| > 1$.
 
The root test works okay too...
 
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...
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