Using ratio test to determine the convergence or divergence of a series

[Nick11234]

Hi all,

I've recently completed the following problem with the aid of a calculator.
The question asks to determine if the series converges or diverges using the ratio test.

I used the limit function in my calculator to check the limit as n approaches infinity of the function derived from the ratio test and got 0, so the function converges.

My question is how can I determine the limit of this function derived from the ratio test by hand, and how would you justify your answer?

 

[JeffM]

You can avoid a lot of opportunity for algebraic error if you supplement the ratio test with some other ideas.

The sum of two convergent series is a convergent series leads to

[math]\dfrac{3^n + \sin(n)}{n!} = \dfrac{3^n}{n!} + \dfrac{\sin(n)}{n!}.[/math]
Now apply the ratio test twice.

[math]\left | \dfrac{3^{(n+1)}}{(n+1)!} \div \dfrac{3^n}{n!} \right | = \left | \dfrac{3^{(n+1)}}{3^n} * \dfrac{n!}{(n+1)!} \right | = \dfrac{3}{n+1}.\\ \lim_{n \rightarrow \infty} \dfrac{3}{n+1} = 0 < 1.[/math]
The next step would be?

Hint:

[math]\alpha \ne 0 \implies \left | \dfrac{\beta}{\alpha} \right | = \dfrac{|\beta|}{|\alpha|}.[/math]

 

[lookagain]

Because \(\displaystyle \ \dfrac{3^n + \sin(n)}{n!} = \dfrac{3^n}{n!} \ + \ \dfrac{\sin(n)}{n!}, \ \ \) and the limit as n --> oo of \(\displaystyle \ \dfrac{\sin(n)}{n!} \ \) readily equals 0 as sin(n)
has a maximum of only 1, then the problem can be simplified to finding the limit as n -->oo of \(\displaystyle \ \dfrac{3^n}{n!} .\)

 

[Deleted member 4993]

Because \(\displaystyle \ \dfrac{3^n + \sin(n)}{n!} = \dfrac{3^n}{n!} \ + \ \dfrac{\sin(n)}{n!}, \ \ \) and the limit as n --> oo of \(\displaystyle \ \dfrac{\sin(n)}{n!} \ \) readily equals 0 as sin(n)
has a maximum of only 1, then the problem can be simplified to finding the limit as n -->oo of \(\displaystyle \ \dfrac{3^n}{n!} .\)

If your computer has a number pad, then you can "type" ∞ with [alt]236

 

[Steven G]

¡ ™ £ ¢ ∞ § ¶ • ª º – ≠ œ ∑ ´ ® † ¥ ¨ ˆ ø π “‘« å ß ∂ ƒ © ˙ ∆ ˚ ¬ … æ Ω ≈ ç √ ∫ ˜ µ ≤ ≥ ÷

This is cool, thanks SK. For the record I have to type alt + 5 to get ∞