Series, Comparison and Limit Tests

[mateomy]

Im practicing problems for my Calc 2 test tomorrow. I am doing this problem which I am not quite sure I've done it right. I think I have, but I want confirmation...

<br /> \sum_{n=1}^{\infty} \frac{1}{\sqrt(n+4)}<br />

I chose my comparison series as \frac{1}{\sqrt(n)} and then ran a limit test eventually finding that it came out to equal 1, which shows that they are behaviorally similar. And since I know my comparison Series is larger than the given Series, I know the latter diverges.

Is that correct?

 

[Mark44]

Im practicing problems for my Calc 2 test tomorrow. I am doing this problem which I am not quite sure I've done it right. I think I have, but I want confirmation...

<br /> \sum_{n=1}^{\infty} \frac{1}{\sqrt(n+4)}<br />

I chose my comparison series as \frac{1}{\sqrt(n)} and then ran a limit test eventually finding that it came out to equal 1, which shows that they are behaviorally similar. And since I know my comparison Series is larger than the given Series, I know the latter diverges.

Is that correct?

You can omit the phrase "And since I know my comparison Series is larger than the given Series".

Since the ratio in the Limit Comparison Test (which is what you used) is 1, and since your comparison series is a divergent series, then your series diverges as well.

The fact that your series is larger than or smaller than (on a per-term basis) the corresponding term of a divergent series, is irrelevant. If you were using the Comparison Test, however, this information would be relevant. In the Comparison Test, for your series to diverge, each term of your series would have to be larger than the corresponding term of a divergent series.

 

[mateomy]

I know, I realized that after I posted it. Is it correct otherwise?

 

[mateomy]

Thanks. (I had to delete a follow up question because I just jumped on a response after skimming over yours.)