could someone help me with the attached problems? Most appreciated.

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- Apr 13th 2010, 10:45 AMazarueA little help with a limit of a series, actuallly two...
could someone help me with the attached problems? Most appreciated.

- Apr 13th 2010, 11:02 AMFailure
The first limit is certainly 0, for consider:

$\displaystyle 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}}$

$\displaystyle =\frac{2n}{n^2+n}=\frac{2}{n+\frac{1}{n}}\rightarr ow 0,\text{ for }n\rightarrow \infty$

the second inequality sign holds true, because from $\displaystyle |\sin(k)|\leq 1$ it follows that $\displaystyle \frac{k}{|\sin(k)|}\geq k$, and thus

$\displaystyle \sum_{k=1}^n \frac{k}{|\sin(k)|}\geq \sum_{k=1}^n k$. - Apr 13th 2010, 11:23 AMtonio

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

that your limit is...

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

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

Tonio - Apr 13th 2010, 03:50 PMazarueThank you for your post.
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).

$\displaystyle

\Longrightarrow \frac{\sum^n_{i=1}\frac{i}{|\sin i|}}{n}\xrightarrow [n\to\infty]{}\infty

$

Could you clarify please ?

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

Thank you for the help. - Apr 13th 2010, 06:05 PMtonio
- Apr 14th 2010, 04:03 AMazaruelimits studies
Tonio, I spent some time reading books on the subject. thank you for you help.