Infinite Series Proof.

**General** · January 15th 2010, 05:50 AM

If $\sum_{n=1}^{\infty} a_n$ diverges and $c\neq0$ Show that :
The series $\sum_{n=1}^{\infty} ca_n$ diverges also.

First I thought about nth terms test, but it failed.
since $\lim_{n\to\infty} a_n$ can be zero.

Then:
I cosidered that sequence of the partial sums for the first seires $S_n=a_1+a_2+a_3+...+a_n$
Since the series diverges, then { $S_n$ } diverges.
i.e. $\lim_{n\to\infty} S_n = \infty$ , $-\infty$ or D.N.E
The sequence of partial sums for the second series , denotes by $(S_n)^{'}$ , is $(S_n)^{'} = cS_n$
Clearly:
$\lim_{n\to\infty} (S_n)^{'} = c \lim_{n\to\infty} S_n$
if $\lim_{n\to\infty} S_n = \pm\infty$ then $\lim_{n\to\infty} (S_n)^{'} = \pm\infty$ depending on the sign of $c$
Hence, The series $\sum_{n=1}^{\infty} ca_n$ diverges.
But what if $\lim_{n\to\infty} S_n$ 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?

**Krizalid** · January 15th 2010, 06:05 AM

$\underset{a\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{a}{ca_{n}}=c\underset{a\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{a}{a_{n}}=c\sum\limits_{n=1 }^{\infty }{a_{n}},$ and the last series was given as divergent, so $\sum ca_n$ diverges as well.

**General** · January 15th 2010, 06:12 AM

Originally Posted by Krizalid

$\underset{a\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{a}{ca_{n}}=c\underset{a\to \infty }{\mathop{\lim }}\,\sum\limits_{n=1}^{a}{a_{n}}=c\sum\limits_{n=1 }^{\infty }{a_{n}},$ and the last series was given as divergent, so $\sum ca_n$ diverges as well.

Thanks.
But what about my proof ?

**Krizalid** · January 15th 2010, 06:14 AM

I see no problem with it.

As for if the limit doesn't exist, it's nothing, that's all.

**General** · January 15th 2010, 06:15 AM

Originally Posted by Krizalid

I see no problem with it.

As for if the limit doesn't exist, it's nothing, that's all.

Thanks.
So the limit which obtained by multiplying a D.N.E limit with a constant is D.N.E also ?

**Krizalid** · January 15th 2010, 06:17 AM

$\lim_{n\to\infty}(-1)^n$ doesn't exist; what about $\lim_{n\to\infty}c(-1)^n,$ the conclusion is the same, it doesn't exist.

**Drexel28** · January 15th 2010, 08:02 AM

Originally Posted by General

If $\sum_{n=1}^{\infty} a_n$ diverges and $c\neq0$ Show that :
The series $\sum_{n=1}^{\infty} ca_n$ diverges also.

First I thought about nth terms test, but it failed.
since $\lim_{n\to\infty} a_n$ can be zero.

Then:
I cosidered that sequence of the partial sums for the first seires $S_n=a_1+a_2+a_3+...+a_n$
Since the series diverges, then { $S_n$ } diverges.
i.e. $\lim_{n\to\infty} S_n = \infty$ , $-\infty$ or D.N.E
The sequence of partial sums for the second series , denotes by $(S_n)^{'}$ , is $(S_n)^{'} = cS_n$
Clearly:
$\lim_{n\to\infty} (S_n)^{'} = c \lim_{n\to\infty} S_n$
if $\lim_{n\to\infty} S_n = \pm\infty$ then $\lim_{n\to\infty} (S_n)^{'} = \pm\infty$ depending on the sign of $c$
Hence, The series $\sum_{n=1}^{\infty} ca_n$ diverges.
But what if $\lim_{n\to\infty} S_n$ 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?

Let $S_m=\sum_{k=1}^{m}a_k$ . Our assumption is that $\lim\text{ }S_m$ does not exist. Assume that $\lim\text{ }cS_m$ existed and equaled $L$ . So, $\lim\text{ }cS_m=L$ . Now, since $c\ne0$ and the left hand limit exists we see that $\lim\text{ }cS_m\cdot\lim\text{ }\frac{1}{c}=\lim\text{ }S_m=\frac{L}{c}$ . Hence, we see that $\lim\text{ }S_m$ exists. Contradiction.

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