infinite series 3

[mario99]

Determine whether the series is convergent or divergent.

\(\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} \frac{3}{k}\)

Any help would be appreciated!

 

[Deleted member 4993]

Determine whether the series is convergent or divergent.

\(\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} \frac{3}{k}\)

Any help would be appreciated!

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem.

 

[mario99]

\(\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} \frac{3}{k} = (-1)^{k+1} \sum_{k=1}^{\infty} \frac{3}{k} \)

It diverges.

 

[Dr.Peterson]

\(\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} \frac{3}{k} = (-1)^{k+1} \sum_{k=1}^{\infty} \frac{3}{k} \)

It diverges.

You can't pull out a factor containing k as you did. The index k is defined only within the summation.

But what does that factor (when kept where it belongs) tell you about the series? If necessary, write out the first few terms explicitly.

 

[nasi112]

Alternating series test is the best choice in this case.

 

[Steven G]

\(\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} \frac{3}{k} = (-1)^{k+1} \sum_{k=1}^{\infty} \frac{3}{k} \)

It diverges.

OK, you pulled out the (-1)^k+1 in front of the summation symbol. My question is what value of k are you planning on using to evaluate (-1)^k+1? If you don't know that answer, then maybe you should not be pulling it out in front the summation symbol.

 

[Steven G]

\(\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} \frac{3}{k} = (-1)^{k+1} \sum_{k=1}^{\infty} \frac{3}{k} \)

It diverges.

Even if what you wrote is correct, you never said why it diverges! Does it diverge by the p-series test with p = ?, does it diverge by the alternating series test and why or does it diverge by the integral test and why, or ....