\sum\limits_{n = 1}^\infty (-1)^n a_n where a_n = f(n) converges if
\sum\limits_{n = 1}^\infty \frac{1}{n^p} converges only if p > 1.
The case p=1 is the harmonic series \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots, which diverges.
\sum\limits_{n = 1}^\infty (-1)^n b_n converges if
\sum\limits_{n = 1}^\infty a_n converges absolutely if \lim_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1.
If the limit equals 1, then the test is inconclusive.
\sum\limits_{n = 1}^\infty a_n converges absolutely if \lim_{n \rightarrow \infty} \sqrt{|a_n|} < 1.
If the limit equals 1, then the test is inconclusive.
To evaluate the series \sum\limits_{n = 1}^\infty a_n, we can define the partial sum S_m = \sum\limits_{n = 1}^m a_n. The value of the series is then the limit of the partial sums S_m, S_\infty = \sum\limits_{n = 1}^\infty a_n = \lim_{m \rightarrow \infty} S_m.
\sum\limits_{n = 1}^\infty (a_n - a_{n-1}), is a telescoping series that evaluates to a_\infty - a_0.
[Don’t memorize the above. You must expand out a few terms of the series to see the pattern, then once you start cancelling, the answer will reveal itself.]
S = \sum\limits_{n = 1}^\infty \alpha r^{n-1} is convergent if |r| < 1. The sum is then S = \frac{\alpha}{1-r}.
The power series is defined as \sum\limits_{n = 0}^\infty c_n (x-a)^n. The geometric series is a special case (when c_n is just a constant, and a=0).
We can use the above geometric series result when representing the function as a power series f\left(x \right) = \frac{c}{{1 - x}} \implies f(x) = \sum\limits_{n = 1}^\infty c x^{n-1}
Power series convergence: there are three possibilities:
Let the coefficient of the n\text{th}-term be c_n. The radius R and interval of convergence, can be found, using the Ratio Test method, as \lim_{n \rightarrow \infty} \left|\frac{c_{n+1}}{c_n}\right| < 1
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n
The “best” approximation of a function by a rational function of given order. Under this technique, the approximant’s power series agrees with the power series of the function it is approximating.
An [m/n] Padé approximant of a function f has the form R_{[m/n]}(x) = \frac{ \sum_{i=0}^{m} a_i x^i }{ 1 + \sum_{i=1}^{n} b_i x^i }, constructed so that R_{[m/n]}(x) = T_{n + m}(x) + O(x^{n+m}), where T(x) is the Taylor approximation of f. The Padé approximant R_{[m/0]} corresponds to the Maclaurin series.
The Padé approximant often gives better approximation of the function than truncating its Taylor series, and it may still work where the Taylor series does not converge. This is since \lim_{x \to \infty} R_{[m/n]}(x) = 0 while \lim_{x\to \infty}T_n(x) = \pm \infty.