Recent content by cpyap

Question edited. Just try to make the question b) clearer. Thanks!

How about b)?

lim_{n\rightarrow\infty}\frac{\frac{2}{3^{n}}+3}{\frac{2n+2}{3^{n}}+n+1} = \frac{3}{\infty+1} = 0 I believe I'm getting tired, missed a whole point here. Thanks a lot for help.

Homework Statement What will be the limit of a) lim_{n\rightarrow\infty}\frac{2+3^{n+1}}{(n+1)(2+3^{n})} b) lim_{x\rightarrow\infty} (x+1)ln(x+1) - x ln(x) - 1 Homework Equations - The Attempt at a Solution It is actually coming from infinite series question, which I goes into the limit step...

Thanks, and thanks for everyone that helps. :biggrin:

So, the final answer should look like this: \sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} n(n+1)(n+2) \geq n3 \frac{1}{n(n+1)(n+2)} \leq \frac{1}{n^{3}} 4n+3 \leq 7n , for all n \geq 1 \frac{4n+3}{n(n+1)(n+2)} \leq \frac{7n}{n^{3}} \sum_{n=1}^{\infty}\frac{7n}{n^{3}} =...

Got it! Thanks!

How do you get the 7n?

Opps, sorry, found the careless mistake. should be n(n+1)(n+2) > n3 so \frac{4n+3}{n(n+1)(n+2)} < \frac{4n+3}{n^{3}} Thanks for pointing me out \frac{4n+2}{n^{3}} = \frac{4}{n^{n}} + \frac{3}{n^{3}} Both converge Therefore, \sum^{\infty}_{n=1} \frac{4n+3}{n(n+1)(n+2)} converges.

Isn't this the same with the n^{3} > n(n+1)(n+1) above?

Thanks for your help. This is what I done: n^{3} > n(n+1)(n+1) \frac{4n+3}{n^{3}} < \frac{4n+2}{n(n+1)(n+2)} \frac{4n+2}{n^{3}} = \frac{4}{n^{n}} + \frac{3}{n^{3}} Both converge Therefore, \sum^{\infty}_{n=1} \frac{4n+3}{n(n+1)(n+2)} converges.Is that correct?

The solution to inhomogeneous equation of L\underline{u}=\underline{g} is the parametric solution of L\underline{u}=\underline{0} + \underline{g}

Homework Statement Test if the infinite series converge or diverge. Homework Equations \sum_{n=1}^{\infty}\frac{4n+3}{n(n+1)(n+2)} The Attempt at a Solution I tried Ratio test: a_{n+1} = \frac{4n+7}{(n+1)(n+2)(n+3)} a_{n} = \frac{4n+3}{n(n+1)(n+2)} \left|\frac{a_{n+1}}{a_{n}}\right| =...