Here is a short review of numerical series which you may find helpful.
REVIEW OF NUMERICAL SERIES
SEQUENCES
A sequence is a list of real numbers. It is called convergent if it has a limit. An increasing sequence has a limit when it has an upper bound. SERIES
(Geometric series,rational numbers as decimals, harmonic series,divergence test)
Given numbers forming a sequence $a_1,a_2,...,$ let us define the nth partial sum as sum of the first n of them $s_n = a_1 +...+ a_n.$
The SERIES is convergent if the SEQUENCE $s_1,s_2,s_3,...$ is. In other words it converges if the partial sums of the series approach a limit.
A necessary condition for the convergence of this SERIES is that a's have limit 0. If this fails, the series diverges.
The harmonic series 1+(1/2)+(1/3)+... diverges.
This illustrates that the terms $a_n$ having limit zero does not guarantee the convergence of a series.
A series with positive terms ,i.e. $a_n > 0$ for all n, converges
exactly when its partial sums have an upper bound.
The geometric series $\displaystyle \sum_{n=1}^{ \infty } r^n$ converges exactly when $-1

INTEGRAL AND COMPARISON TESTS
(Integral test,p-series, comparison tests for convergence and divergence, limit comparison test)

Integral test: Suppose $f(x)$ is positive and DECREASING for all large enough x. Then the following are equivalent:
I. $\displaystyle \int_1^{ \infty } f(x)dx$ is finite, i.e. converges.
S. $\displaystyle \sum_{n=1}^{ \infty } \, f(n)$ is finite, i.e. converges.
This gives the p - test: $\displaystyle \sum_{n=1}^{ \infty } \frac{1}{n^p}$ converges exactly when $p > 1 .$

Comparison test: Suppose there is a fixed number K such that
for all sufficiently large n: $0 < a_n < K b_n .$
Convergence. If $\displaystyle \sum_{n=1}^\infty b_n$ converges then so does $\displaystyle \sum_{n=1}^\infty a_n.$
Divergence. If $\displaystyle \sum_{n=1}^\infty a_n$diverges then so does $\displaystyle \sum_{n=1}^\infty b_n$.
(Positive series having smaller terms are more likely to converge.)
Limit comparison test: SUPPOSE: $a_n > 0$, $b_n > 0$ and
$\displaystyle \lim_{n \to \infty } \frac{a_n}{b_n} = R$ exists. Moreover, R is not zero.
THEN $\displaystyle \sum_{n=1}^\infty a_n$ and $\displaystyle \sum_{n=1}^\infty b_n$
both converge or both diverge.
OTHER CONVERGENCE TESTS FOR SERIES
(Alternating series test, absolute convergence, RATIO TEST)
Alternating series test: Suppose the sequence $a_1,a_2,a_3,...$ is decreasing and has limit zero. Then $\displaystyle \sum_{n=1}^\infty {(-1)}^n a_n$ converges.
This applies to (1)-(1/2)+(1/3)-(1/4)+...

Absolute Convergence Test: IF $\displaystyle \sum_{n=1}^\infty \vert a_n \vert$ converges,
THEN $\displaystyle \sum_{n=1}^\infty a_n$ converges.
Ratio test:
SUPPOSE $\vert { \frac {a_{n+1}}{a_n}} \vert$ has limit equal to r.
IF $r < 1$ then $\displaystyle \sum_{n=1}^{\infty} a_n$ CONVERGES.
IF $r > 1$ the $\displaystyle \sum_{n=1}^{\infty} a_n$ DIVERGES.