# A little help with a limit of a series, actuallly two...

• Apr 13th 2010, 11:45 AM
azarue
A little help with a limit of a series, actuallly two...
could someone help me with the attached problems? Most appreciated.
• Apr 13th 2010, 12:02 PM
Failure
Quote:

Originally Posted by azarue
could someone help me with the attached problems? Most appreciated.

The first limit is certainly 0, for consider:

$0\leq \frac{n}{\sum_{k=1}^n\frac{k}{|\sin(k)|}}
\leq \frac{n}{\sum_{k=1}^n k}=\frac{n}{\frac{n(n+1)}{2}}$

$=\frac{2n}{n^2+n}=\frac{2}{n+\frac{1}{n}}\rightarr ow 0,\text{ for }n\rightarrow \infty$
the second inequality sign holds true, because from $|\sin(k)|\leq 1$ it follows that $\frac{k}{|\sin(k)|}\geq k$, and thus
$\sum_{k=1}^n \frac{k}{|\sin(k)|}\geq \sum_{k=1}^n k$.
• Apr 13th 2010, 12:23 PM
tonio
Quote:

Originally Posted by azarue
could someone help me with the attached problems? Most appreciated.

Since $0<|\sin n|<1\,\,\,\forall\,n\in\mathbb{N}$ (why? Hint: $\pi$ is irrational), we get that $\frac{n}{|\sin n|}>n\Longrightarrow \frac{n}{|\sin n|}\xrightarrow [n\to\infty]{}\infty$ $\Longrightarrow \frac{\sum^n_{i=1}\frac{i}{|\sin i|}}{n}\xrightarrow [n\to\infty]{}\infty$ as well, and from here

As for the second one: $\left(1-\frac{2}{n+2}\right)^{n+2}\xrightarrow [n\to\infty]{}e^{-2}\Longrightarrow$ the limit is...

Theorems used: if $0\leq a_n\xrightarrow [n\to\infty]{}a$ then $\frac{a_1+a_2+\ldots+a_n}{n}\xrightarrow [n\to\infty]{}a$ and also $\sqrt[n]{a_1\cdot a_2\cdot\ldots\cdot a_n}\xrightarrow [n\to\infty]{}a$ , with the obvious generalizations when $a=\infty$

Tonio
• Apr 13th 2010, 04:50 PM
azarue
tonio,

in the original question we had to compute a limit who looked something like this: (n)/(sum of something). in your solution you took the opposite, (some of something/n).
$
\Longrightarrow \frac{\sum^n_{i=1}\frac{i}{|\sin i|}}{n}\xrightarrow [n\to\infty]{}\infty
$

In addition, may I ask for more detailed solution for question number 2.

Thank you for the help.
• Apr 13th 2010, 07:05 PM
tonio
Quote:

Originally Posted by azarue
tonio,

in the original question we had to compute a limit who looked something like this: (n)/(sum of something). in your solution you took the opposite, (some of something/n).
$
\Longrightarrow \frac{\sum^n_{i=1}\frac{i}{|\sin i|}}{n}\xrightarrow [n\to\infty]{}\infty
$

This is the generalization I talked about in my post: you figure this (easy) one out and deduce your limit is zero

In addition, may I ask for more detailed solution for question number 2.

Look at the geometric mean sequence that follows from the one in my post. You can also try to google "arithmetic and geometric means limits".
For example, look at Limit of arithmetic mean , or here More Challenging Limits ( the second one contains proofs of both theorems I quote)

Tonio

Thank you for the help.

.
• Apr 14th 2010, 05:03 AM
azarue
limits studies
Tonio, I spent some time reading books on the subject. thank you for you help.