Suppose (a_{n}) is a monotonically decreasing sequence of positive numbers, and that $\displaystyle \Sigma\limits_{n=1}^\infty a_{n}$ converges.

Show that $\displaystyle \lim_{n\rightarrow\infty} na_{n} = 0$

Printable View

- Mar 17th 2013, 07:37 AMKanwar245Monotonically Decreasing Sequence
Suppose (a

_{n}) is a monotonically decreasing sequence of positive numbers, and that $\displaystyle \Sigma\limits_{n=1}^\infty a_{n}$ converges.

Show that $\displaystyle \lim_{n\rightarrow\infty} na_{n} = 0$ - Mar 17th 2013, 10:01 AMMINOANMANRe: Monotonically Decreasing Sequence
This is a well known theorem in the series of positive terms.

If the series Σan of positive monotone decreasing terms is to converge then we must have not only lim(an)=0 but also lim(n(an))=0 . However the condition lim(n(an))=0 is only necessary , not a sufficient one for these type of series. If lim(nan) does not tend to zero then definitely the series diverges but lim(nan)=0 does not necessarily implies anything as to the possible convergence of the series. in fact the Abel series Σ(1/nlogn) diverges though lim(nan)=0

Get a good book on series to revise all these theorems.

MINOAS - Mar 18th 2013, 04:39 PMKanwar245Re: Monotonically Decreasing Sequence
Can't I use the fact that since $\displaystyle \Sigma\limits_{n=1}^\infty a_{n}$ converges implies $\displaystyle \lim_{n\rightarrow\infty}a_{n} = 0$

So, $\displaystyle \lim_{n\rightarrow\infty}na_{n} = \lim_{n\rightarrow\infty}n\lim_{n\rightarrow\infty }a_{n} = \lim_{n\rightarrow\infty}n * 0 = 0$ - Mar 18th 2013, 08:18 PMhollywoodRe: Monotonically Decreasing Sequence
No, that definitely doesn't work.

But you know (by definition) that $\displaystyle \sum_{n=1}^\infty{a_n} = \lim_{n\to\infty}\sum_{k=1}^n{a_k}$. So you just need to show that the finite sum is less than $\displaystyle na_n$.

- Hollywood