Question

If \(\sum {{a_n}} \)is divergent and \(c \ne 0\), show that \(\sum {c{a_n}} \)is divergent

Short answer

\(\sum {c{a_n}} \)is divergent

Step-by-step solution

Definition of Series and Divergent Series

A Series is known as the sum of infinite sequence of numbers. A Series is said

to be divergent only if the sequence of the partial sum does not reach a limit and

it will be usually∞

Given parameters

The given Series is \(\sum {{a_n}} \)and the series is a divergentseries. The value of \(c\)is a non

zero constant. Then we need to prove that \(\sum {c{a_n}} \)is alsodivergent.

Comparing the given series with the infinite series convergence along with partial sums

Let us take the definition of infinite series convergence with partial sums.

\({S_k}\)is the \({k^{th}}\)partial sum of the series\(\sum\limits_{n = 1}^\infty {c{a_n}} \)

We can rewrite the above definition as

\(\begin{aligned}{s_k} = ca1 + ca2 + ca3 + .....cak\\ {s_k} = c\left( {a1 + a2 + a3 + .....ak} \right)\\\end{aligned}\) -- Equation 1

Letusapply the limits of \({S_k}\) on both left and right sides to the Equation 1

\(\mathop {\lim }\limits_{k = \infty } {s_k} = \mathop {\lim }\limits_{k = \infty } c\left( {a1 + a2 + a3 + .....ak} \right)\)-- Equation 2

Proving \(\sum {c{a_n}} \)is a divergent series and not a convergent series

In case \(\sum {{a_n}} \)is convergent then \(\mathop {\lim }\limits_{k = \infty } c\left( {a1 + a2 + a3 + .....ak} \right)\)will be also convergent

series and c is a non-zero constant number.

Let us multiply \(\frac{1}{c}\) to the equation \(\mathop {\lim }\limits_{k = \infty } c\left( {a1 + a2 + a3 + .....ak} \right)\)

\(\mathop {\lim }\limits_{k = \infty } \frac{1}{c}.c\left( {a1 + a2 + a3 + .....ak} \right)\)

\( = \mathop {\lim }\limits_{k = \infty } \left( {a1 + a2 + a3 + .....ak} \right)\)

Hence \(\mathop {\lim }\limits_{k = \infty } \left( {a1 + a2 + a3 + .....ak} \right)\)is the sequence of partial sum of the sequence\(\sum {{a_n}} \)and we can say \(\sum {{a_n}} \)is a convergent series.

But we started the assumption as divergent series and hence \(\sum {{a_n}} \)is not a convergent series.

Thus\(\sum {c{a_n}} \)cannot be a convergent series. It is a divergent series.