series 1

Calculus Level 2

Test the series: $\displaystyle\frac{1}{2}+\displaystyle\frac{2}{3}+\displaystyle\frac{3}{4}+\displaystyle\frac{4}{5}+\displaystyle\cdot\cdot\cdot$

inconclusive it diverges it converges neither of the answers above

2 solutions

Daniel Liu
Mar 8, 2015

Note that each term is $\ge \dfrac{1}{2}$ , so the sum $S > \dfrac{1}{2}+\dfrac{1}{2}+\cdots$ which diverges, so the sum $S$ also diverges.

Another way-

$= 1 - \dfrac{1}{2} + 1 - \dfrac{1}{3} + 1 - \dfrac{1}{4} + .......$

$\displaystyle\lim_{n \to \infty} \sum_{1}^{n}\left( n - \dfrac{1}{n}\right)$

U Z - 6 years, 3 months ago

Jon Haussmann
Mar 9, 2015

The series fails even the most basic nth term test .

It fails? how?

samuel ayinde - 6 years, 3 months ago

If a series converges, then the nth term must go to 0 - this is the nth term test. In your series, $\lim_{n \to \infty} \frac{n}{n + 1} = 1,$ so it fails the nth term test.

Jon Haussmann - 6 years, 3 months ago

Isn't the nth term test only indicative of divergence? It's only conclusive if the limit goes to zero, after which we can say that the sum diverges.

Zach Abueg - 3 years, 10 months ago

@Zach Abueg – I don't understand what you're saying.

If a series converges, then the nth term must go to 0. The contrapositive then tells us that if the nth term does not go to 0, then the series must diverge, which is then the answer to this problem.

Jon Haussmann - 3 years, 10 months ago

@Jon Haussmann – Sorry, I meant if the nth term doesn't go to 0, then it must diverge. I originally thought you were saying that if the nth term went to 0, then it must converge, but is not necessarily true, but I just misunderstood. A little misunderstanding on my part, as well - I get your comment now :)

Zach Abueg - 3 years, 10 months ago

series 1

2 solutions

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