A. $\sum_\limits{n=1}^\infty \frac{1}{n}$
B. $\sum_\limits{n=0}^\infty (1.5)^n$
C. $\sum_\limits{n=0}^\infty \frac{4n-1}{5n+1}$
D. $\sum_\limits{n=1}^\infty (e)^{-2n}$
So my thinking so far is that everything diverges except D, but I don't know how to specifically show it. The reason I believe so is because as n increases $(1/e^{-2})^n$ will get extremely close to zero meaning that soon you will basically be adding 0 so it will converge. However I don't know if that is correct, or if it is the correct way to think about it. Is there a convergence test that will help me in this case?
$\endgroup$ 32 Answers
$\begingroup$Hint. One may recall that each series that converges has its general term tending to $0$, here $$ \lim_{n \to \infty}(1.5)^n=? \qquad \lim_{n \to \infty}\frac{4n-1}{5n+1}=? $$ One may recall that the geometric series $\displaystyle \sum_{n=0}^\infty x^n$ is convergent if and only if $|x|<1$. Concerning $A$, one may recall the nature of the harmonic series.
$\endgroup$ $\begingroup$A diverges as harmonic
B divergent as geometric$(1.5>1)$
C the general term doesn't go to zero
D converges as geometric$(e^{-2}<1)$.
