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**General** If $\displaystyle \sum_{n=1}^{\infty} a_n$ diverges and $\displaystyle c\neq0$ Show that :

The series $\displaystyle \sum_{n=1}^{\infty} ca_n$ diverges also.

First I thought about nth terms test, but it failed.

since $\displaystyle \lim_{n\to\infty} a_n$ can be zero.

Then:

I cosidered that sequence of the partial sums for the first seires $\displaystyle S_n=a_1+a_2+a_3+...+a_n$

Since the series diverges, then { $\displaystyle S_n$ } diverges.

i.e. $\displaystyle \lim_{n\to\infty} S_n = \infty$ , $\displaystyle -\infty$ or D.N.E

The sequence of partial sums for the second series , denotes by $\displaystyle (S_n)^{'}$, is $\displaystyle (S_n)^{'} = cS_n$

Clearly:

$\displaystyle \lim_{n\to\infty} (S_n)^{'} = c \lim_{n\to\infty} S_n$

if $\displaystyle \lim_{n\to\infty} S_n = \pm\infty$ then $\displaystyle \lim_{n\to\infty} (S_n)^{'} = \pm\infty$ depending on the sign of $\displaystyle c$

Hence, The series $\displaystyle \sum_{n=1}^{\infty} ca_n$ diverges.

But what if $\displaystyle \lim_{n\to\infty} S_n$ D.N.E ?

i.e. If you multiply a D.N.E Limit with by a constant, It will be D.N.E too ?

My proof is right?