0
$\begingroup$

"show that the given series $(3/2-4/3)+(5/4-6/5)+....... $ is convergent. Give reason for obtaining different nature of the series before and after grouping the terms." The nth term of this series is $(-1)^{n+1} \dfrac{n+2}{n+1}$. I approach by Leibnitz's test but this gives the series is divergent, please help me.

  • 0
    You may need to check the problem sheet again. The reason you fail in showing convergence is that this series diverges. Try proving that instead. – 2017-02-21
  • 1
    But i need showing convergent because question says show convergent. – 2017-02-21
  • 1
    I have done some correction of this question. – 2017-02-21
  • 2
    You don't show the series is convergent because the question says so. You show convergence if it **does** converge, otherwise it does not, no matter how badly one may want it to. Now, if the $n$'th term is $(-1)^{n+1}(n+2)\over n+1$, then the series diverges. If, however, the $n$'th term is understood as _the thing in brackets_, then it's another story (and another formula, BTW). – 2017-02-21
  • 0
    You may already have been told that if a term is of the form $\frac{(-1)^{n+1}(n+2)}{n+1}$, then the series diverges. However, I think the expression in brackets corresponds to one term of the series. So for example, your first term is $\frac 32 - \frac 43 = \frac 16$, your second term is $\frac 54 - \frac 65 = \frac 1{20}$, so your series is $\frac 16 + \frac{1}{20} + \ldots$. Please tell me if this is correct, then I can tell you this series is convergent, and prove it in an answer. But I need confirmation. – 2017-02-21
  • 1
    @ actoh.. i can't make the nth term of 6,20,42,72.... – 2017-02-21
  • 1
    @actoh.. why you told 1/6+1/20+1/42+1/72 is convergent? – 2017-02-21
  • 0
    Because, for example, you may compare it with $1\over n^2$ which is known to be convergent. But still, you have to derive the n'th term first. – 2017-02-21
  • 1
    But i can't make nth term. – 2017-02-21
  • 0
    You already did, in a way. Now you split all those $(-1)^{n+1}(n+2)\over n+1$ into pairs and find the value of one such pair. – 2017-02-21

2 Answers 2

7

The problem is that your nth term is incorrect.

$$\frac{2n+1}{2n} - \frac{2n+2}{2n+1} $$ Will get you $(3/2 - 4/3)$ for $n=1$, $(5/4 - 6/5)$ for $n=2$ and so on.

Simplifying, this makes the nth term: $$\frac{(2n+1)^2 - 2n(2n+2)}{2n(2n+1)} $$ $$=\frac{4n^2 + 4n + 1 - 4n^2 - 4n}{2n(2n+1)} $$ $$=\frac {1}{2n(2n+1)} $$

Now, $$\frac{1}{4n^2 + 2n} < \frac{1}{n^2}$$ for all natural numbers $n $, and it is well-known (and fairly straightforward to prove) that the sum of $\frac{1}{n^2} $ converges (to $\frac{\pi^2}{6}$ in fact), so that we can say the sum in question is bounded above by $\pi^2 /6$ and bounded below by $1/6$

With some clever manipulation, we can find the exact value. Expanding $\frac{1}{2n(2n+1)} = \frac{1}{2n} - \frac{1}{2n+1} $: $$1/2 - 1/3 + 1/4 - 1/5 +.... $$ $$ = -(-1/2 + 1/3 - 1/4 + 1/5 -...) $$ $$= 1 - (1 - 1/2 + 1/3 - 1/4 + 1/5 - ...) $$ The sum on the right is just the alternating harmonic series, which converges to $\ln(2)$ (which can be seen by using the Taylor expansion of $\ln(1+x)$). Therefore, the original sum converges to $1 - \ln(2)$.

1

If you consider the given series (without grouping terms) then the generic term $a_n$ for stage $n$ is indeed: $a_n=(-1)^{n+1}\frac{n+2}{n+1}$ and hence the series do no converge since $\lim_{n\to\infty}a_n\ne 0$.

For the second part look at the solution of @infinitylord

  • 0
    yes I saw it... going to fix – 2017-02-21
  • 0
    This is incorrect. You only get the desired grouped terms for $n \ge 3$ where $n$ is odd. For example, expanding out $n=4$, you get $\frac{4}{3} - {5}{3}$, which is non-existent in the series. See my solution. – 2017-02-21
  • 0
    @infinitylord I've already looked at... I wanted just to see if there was a way to use telescoping... but maybe not – 2017-02-21
  • 0
    No I don't see any telescoping possibility... so canceling – 2017-02-21
  • 0
    I don't believe the exact value can be obtained by telescoping, but I have updated my answer to include the exact value by means of comparison with the alternating harmonic series. – 2017-02-21
  • 1
    @infinitylord That was the idea but I didn't recall about $\ln(1+x)$ series. Good occasion to revise ;-) – 2017-02-21