1. Series

Series

Given a sequence of real numbers a_n, we can consider a series \sum_{n=1}^{\infty} a_n, simply as a sequence of partial sums:

    \[S_n=a_1+\ldots +a_n\]

The sum of a given series is the limit of sequence S_n, if it exists. If it is finite, then we say that the series converges.

E.g. series \sum_{n=1}^{\infty}\frac{1}{2^n} is convergent and sums up to 2. Indeed, S_n=1+\frac{1}{2}+\ldots \frac{1}{2^n}=2-\frac{1}{2^n}, which converges to 2.

Necessary condition

It is easy to notice that if a series \sum_{n=1}^{\infty} a_n is convergent then the sequence a_n is convergent to zero. In particular, if a_n is not convergent or converges to a non-zero limit, then \sum_{n=1}^{\infty} a_n cannot be convergent.

E.g. series \sum_{n=1}^{\infty} (-1)^n is not convergent, because sequence (-1)^n is not convergent. Also \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^n is not convergent, because \left(1+\frac{1}{n}\right)^n converges to e\neq 0.

Notice that the reverse implication is not true. A series \sum_{n=1}^{\infty} a_n may not be convergent even if a_n converges to zero. E.g. sequence a_n: 1,\frac{1}{2},\frac{1}{2},\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \ldots.

Arithmetic and comparison criterion

Since series can be seen as sequences, many theorems on arithmetic of limits make sense also in the case of series. For example, limit of sum of two series is the sum of their sums. Notice that multiplying is a bit tricky. Multiplication of elements of series is not the same as multiplication of partial sums.

But we have a criterion which is implied by the three sequences theorem. If |a_n|\leq b_n for all n greater then some number n_0 and \sum_{n=1}^{\infty}b_n is convergent then also \sum_{n=1}^{\infty}a_n converges.

E.g. series \sum_{n=1}^\infty \frac{1}{3^n} is obviously convergent, because 0\leq \frac{1}{3^n}\leq \frac{1}{2^n}.

D’Alembert criterion

D’Alembert criterion states what can be said how convergence of \sum_{n=1}^\infty a_n depending on limit of \frac{|a_{n+1}|}{|a_n|}. If this limit is <1, the series is convergent, but if >1, it is not.

E.g. \sum_{n=1}^{\infty}\frac{2^n}{n!} is convergent, because \frac{a_{n+1}}{a_n}=\frac{2}{n+1}\to 0<1.

Notice that the criterion does not give any result if the limit equals 1.

Cauchy criterion

Cauchy criterion states what can be said how convergence of \sum_{n=1}^\infty a_n depending on limit of \sqrt[n]{a_n}. If this limit is <1, the series is convergent, but if >1, it is not.

E.g. \sum_{n=1}^{\infty}\frac{n}{2^n} is convergent, because \sqrt[n]{a_n}=\frac{\sqrt[n]{n}}{2}\to \frac{1}{2}<1.

Notice that the criterion does not give any result if the limit equals 1.

Leibniz theorem

Leibniz theorem states that if a_n is non-increasing and converges to zero, then \sum_{n=1}^\infty (-1)^{n+1} a_n is convergent.

In particular, \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} is convergent, since \frac{1}{n} is non-increasing and converges to zero.

Absolute convergence

We shall say that \sum_{n=1}^\infty  a_n is absolutely convergent, if \sum_{n=1}^\infty  |a_n| is convergent. Notice that, if a series is absolutely convergent, then it is convergent.

E.g. \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} is convergent, but is not absolutely convergent. On the other hand, \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2^n} is absolutely convergent (and also obviously convergent).