# Series of Positive Terms

We shall denote by $${S^ + }$$ the set of all such series whose terms are positive. Thus, $$\sum {u_n}$$ is a series of positive terms if and only if $$\sum {u_n} \in {S^ + }$$. If $$\sum {u_n} \in {S^ + }$$, then its sequence or partial sums $$\left\langle {{s_n}} \right\rangle $$ is monotonically increasing for $${s_{n + 1}} – {s_n} = {u_{n + 1}} > 0$$ $$\forall n$$.

This gives:

** Theorem:** If $$\sum {u_n} \in {S^ + }$$ then $$\sum {u_n}$$ either converges to positive number or diverges to $$ + \infty $$.

** Proof:** If $$\sum {u_n} \in {S^ + }$$ and its sequence of partial sums $$\left\langle {{s_n}} \right\rangle $$ be bounded, $$\exists {\text{ }}k > 0$$ such that $$k > {s_{n + 1}} > {s_n} \geqslant {u_1}{\text{ }}\forall n$$. Thus $$\left\langle {{s_n}} \right\rangle $$ being monotonic and bounded is convergent and $$k \geqslant \lim {s_n} \geqslant {u_1} > 0$$. If $$\left\langle {{s_n}} \right\rangle $$ is bounded, then $$\left\langle {{s_n}} \right\rangle $$ being monotonic for $${s_{n + 1}} – {s_n} = {u_{n + 1}} > 0{\text{ }}\forall n$$ diverges to $$ + \infty $$, i.e. $$\sum {u_n}$$ diverges to $$ + \infty $$. Hence the theorem is proved.

When the direct investigation regarding the boundedness of $$\left\langle {{s_n}} \right\rangle $$ is not easy then we use alternative tests to establish the convergence of the series. Establishing various tests is given as

- If $$\lim {a_n} = a$$ and $${k_1} < a{\text{ }}\exists m \in \mathbb{N}$$ such that $${a_n} > {k_1}{\text{ }}\forall n \geqslant m$$.
- If $$\lim {a_n} = a$$ and $${k_2} < a{\text{ }}\exists m \in \mathbb{N}$$ such that $${a_n} > {k_2}{\text{ }}\forall n \geqslant m$$.

The above two results follow from the fact that all except a finite number of terms of the sequence $${a_n}$$ lie in the neighborhood $$\left( {{k_1},{k_2}} \right)$$ of the limit ‘*a*’.