# Series of Positive Terms

We shall denoting by ${S^ + }$, the set of all such series whose terms are positive. Thus, $\sum {u_n}$ is 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 $\left\langle {{s_n}} \right\rangle$ be 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 proved.

When the direct investigation regarding the boundedness of $\left\langle {{s_n}} \right\rangle$ are not easy then we use alternative tests to establish the convergence of the series. In establishing various tests, 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 follows 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’.