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$$\lim_{n\to\infty}\frac{1}{n+2}-\frac{1}{2(n+1)}$$

Having difficulty with the limit. Would this be solveable, by writing all terms individually, cancelling out; then attempting to find a convinient form?

I am particularly interested in whether this converges...

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1  
Are you familiar with the algebra of limits? –  Adam Rubinson Sep 6 '12 at 23:40
    
i am aware of basic series, geometric and the like only. also i have some understanding that the harmonic series diverges, and that possibly the difference may be convergent. –  mello Sep 6 '12 at 23:47
    
It's very easy to show $0\le \frac{1}{n+2}-\frac{0.5}{n+1} \le \frac{1}{n+2}$ for all $n\ge 0$, and since the right-hand side of this squeeze goes toward $0$... –  Henning Makholm Sep 6 '12 at 23:48
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ah thankyou. somehow i was reading the lim as if it were sigma the sum! (facepalm) much easier as a limmit. thanks –  mello Sep 6 '12 at 23:49
    
sun.iwu.edu/~lstout/limitTheorems/node3.html Read Theorem 3.1 (but don't read the proof... it's probably too advanced for you.) The "a" they use means a real number, but it can also be +infinity or -infinity, as in your example. –  Adam Rubinson Sep 6 '12 at 23:51

2 Answers 2

You could observe that $$ \frac{1}{n+1}-\frac{1}{2(n+1)}=\frac{n}{2(n+2)(n+1)} $$ and take it from there.

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In addition to the answers received, another approach is to write the series expansion at n = Infinity.

You would have something like:

http://www.wolframalpha.com/input/?i=Limt%5B1%2F(n%2B2)%20-%201%2F(2(n%2B1))%2C%20n%20-%3E%20Infinity%5D&t=crmtb01

HTH ~A

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