Steply

Does Σ (-1)^(n+1)/n converge

Step by step, with the reason for every step · Alternating series · Calculus 2

Step by step

The question

Question: decide if the series converges: the sum of (−1)ⁿ⁺¹ over n for n from 1 to infinity.

Σn = 1∞(−1)n+1n =

1 The size

Write the first terms: the signs take turns.

terms = 1, −0.5, 0.333, −0.25, ...

Take the size of each term, without the sign: b n = 1 over n.

bn = 1n

The alternating series test: if the sizes get smaller and smaller and go to 0, the series converges.

alternating series test:
if bn gets smaller and bn → 0,
the series converges

2 Getting smaller?

n + 1 is bigger than n, so the bottom gets bigger: each size is smaller than the one before.

bn + 1 < bn

3 Going to 0?

The bottom grows without end: the sizes go to 0.

as n → ∞ = 1n → 0

4 Decide

Both checks pass: the series converges, by the alternating series test.

= converges

5 Absolutely or conditionally?

Now drop the signs: the sum of 1/n is a p-series with p = 1.

p is not bigger than 1: it diverges.

Σn = 1∞1n : p-series, p = 1

The series converges only because the signs take turns.

That is called converging conditionally.

only with the signs: conditionally

So the series converges conditionally.

= converges conditionally

Careful!

The signs make a difference!

Without them it diverges, but with them taking turns, it converges.

Careful: not Σ 1n diverges, so this diverges ✗

6 Check

Let's check with a calculator: the sum of 100 terms is about 0.6882, of 101 terms about 0.6981.

They jump less and less, and the sum is between them.

It settles.

Right!

S100 ≈ 0.6882, S101 ≈ 0.6981

Let's look back.

First: the size.

Then: getting smaller?.

Then: going to 0?.

Then: decide.

Then: absolutely or conditionally?.

Last: check.

Answer
= converges conditionally

Your turn

Try this one yourself, then check:

Σn = 1∞(−1)n+14√n
Show the answer
converges conditionally

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