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Find the general term of the sequence, starting with n=1, determine whether the sequence converges, and if so find its limit.

$$\frac{3}{2^2 - 1^2}, \frac{4}{3^2 - 2^2} , \frac{5}{4^2 - 3^2}, \cdots$$

Can you help me with this,I know how to solve the problem with n= 0 but is that different with n= 1?

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Observe the series first. $$ \frac{1\color{green}{+2}}{(1\color{green}{+1})^2-(1)^2}, \frac{2\color{green}{+2}}{(2\color{green}{+1})^2-(2)^2}, \frac{3\color{green}{+2}}{(3\color{green}{+1})^2-(3)^2}, .... \text{upto} \frac{n\color{green}{+2}}{(n\color{green}{+1})^2-(n)^2} $$

So now, General Term or $ n^{th} $ term can be written as:

$$ \frac{(n+2)}{(n+1)^2-n^2} $$

$$ \frac{(n+2)}{(\require{cancel} \cancel{n^2}+2n+1- \cancel{n^2}) }$$

$$ = \frac{(n+2)}{(2n+1)}$$

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  • $\begingroup$ thank you for a very detailed answer ! I got everything $\endgroup$ – Hà Nhị Dec 15 '17 at 15:25
  • $\begingroup$ @Hà Nhi I feel delighted to help you! Welcome :) $\endgroup$ – Ravi Prakash Dec 15 '17 at 15:27
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Note that the general formula for the $n$th term is: $$T_n = \frac{n+2}{(n+1)^2-n^2}$$ which can be simplified using difference of squares ($a^2-b^2=(a-b)(a+b)$) as: $$T_n = \frac{n+2}{2n+1}$$

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  • $\begingroup$ I understand your answer but can you give me more information about how to deal with this kind of problem ? I've read on textbook but i feel overwhelmed and can't figure out a proper way $\endgroup$ – Hà Nhị Dec 15 '17 at 14:58
  • $\begingroup$ @HàNhị Please tell on what you want more information. $\endgroup$ – Rohan Dec 15 '17 at 14:59
  • $\begingroup$ Oh I've just read some documentaries and got them...Thank you a lot ! $\endgroup$ – Hà Nhị Dec 15 '17 at 15:11
  • $\begingroup$ At first, I thought it was an unresolved tension. $\endgroup$ – Eric Duminil Dec 15 '17 at 22:35
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General term:

$$\frac{n+2}{(n+1)^2-n^2}$$

That is, $$\frac{n+2}{2n+1}$$

Can you find the limit now?

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The expressions above proves divergence by the nth term test, i.e. the limit of the sequence does not go to zero.

Here's a small survey of other techniques.

Is interesting to note that this sequence leads to an inconclusive Ratio Test, i.e. |T_n+1/T_n|=1.

Let (n+2)/(2*n+1)=T_n

Then 2*T_n = (2*n+4)/(2*n+1) = 1 + 3/(2*n+1).

So, T_n = (1/2) + 3/[2*(2*n+1)].

This expression has a somewhat easier limit to evaluate, however to get there required extra steps. From here though, its easier to approximate the sum.

The constant term, (1/2), guarantees each additional term in the sum adds a half, so part of the Partial Sum is (N/2).

The second term is (3/2)/(2*n+1)= (3/4)/(n+1/2)> (3/4)/(n+1),

so sum(T_n) > (N/2) + (3/4)*sum(1/(k+1), k, 1,N), with the right hand side being a lower bound for the sum.

The second sum is approximately the natural log of N.

So sum(T_n,1,n) > (N/2) + (3/4)*ln(N/e).

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  • $\begingroup$ Welcome to MSE. Please use MathJax to format your answer. $\endgroup$ – user507623 Dec 15 '17 at 18:09

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