I have the following alternating series $1+1/2+1/2-1/2^2+1/3-1/2^3+1/4-1/2^4$ does the alternating series converge or diverge?does the alternating series apply?

I know that the series can be written as $1/n-1/2^n$ which as I can see not alternating, so the alternating test is not applicable $1/n>1/2^n$for all n, but how can I prove that it converges.

  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. It will help clear up this question. $\endgroup$ Jul 16, 2017 at 22:22
  • $\begingroup$ Thanks. I've just made a few changes, and I'll read those posts $\endgroup$
    – Jasmine
    Jul 16, 2017 at 22:27
  • $\begingroup$ The series is $\sum \left( \frac 1n - \frac{1}{2^n} \right) = \sum \frac 1n - \sum \frac{1}{2^n}$. We know that $\sum \frac 1n = \infty$ and $\sum \frac{1}{2^n} = 2$. As $\infty - 2 = \infty$, your series diverge. But you have to write this more mathematically. $\endgroup$ Jul 16, 2017 at 22:33
  • $\begingroup$ 'you have to write this more mathematically' meaning it's not rigorous or even valid? you know convergent sums are not necessarily commutative. Although this might be an exception as the second term converges? $\endgroup$ Jul 16, 2017 at 22:37
  • $\begingroup$ Well, partial sums will diverge, because $\sum \frac 1n$ diverges. That is, for every $M \in \Bbb N$, there exists $N$ such that $\sum^N \frac 1n >M$. This implies that $\sum^N \left( \frac 1n - \frac{1}{2^n} \right)>\text{ (some calculations here) } > M-2$. Hence the series diverges. $\endgroup$ Jul 16, 2017 at 22:41

3 Answers 3


$$u_n=\frac {1}{n}-e^{-n\ln (2)} $$

$$=\frac {1}{n}(1-ne^{-n\ln (2)}) $$ $$\sim \frac {1}{n} \;\;(n\to +\infty) $$

because $\lim_{+\infty}ne^{-n\ln (2)}=0$.

hence, it diverges.


$\sum \frac{1}{n} $ diverges

$\sum \frac {-1}{2^n} $ converges

their sum is divergent.

  • $\begingroup$ I'll try this one, thanks. Am I right that this sequence is not alternating? $\endgroup$
    – Jasmine
    Jul 16, 2017 at 22:29
  • $\begingroup$ well, it's clearly alternating but you can't apply leibniz' rule because the terms are not strictly decreasing. But it's alternating and hence not commutative. It is divergent so it isn't necccessarily associative either. $\endgroup$ Jul 16, 2017 at 22:32

You can try limit comparison test.

$$\lim_{n \to \infty} \frac{\frac 1n - \frac{1}{2^n}}{\frac 1n} = 1$$

Since this limit is a nonzero constant, both series converge or both series diverge.

As $\sum \frac 1n$ diverges by $p-$test, the series $\sum \left( \frac 1n - \frac{1}{2^n} \right)$ also diverges.


Equivalents are very powerful in estimating the behaviour of a series and often ignored.

$u_n>0$ and $\displaystyle u_n=\frac 1n-\frac 1{2^n}\sim\frac 1n$ which is a term of a divergent series so $\sum u_n$ diverges as well.


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