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’.



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