SoonerTheBetter said:
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I am trying to figure out whether a sequence is convergent or not by writing out the first 5 terms. The sequence is: sin[1+(pi/n)]+nsin(pi/n).
What quasar987 meant is to change n to 1 / (1 / n). Since you need a fraction, and you only have a multiplication, so what you should do is to change multiplication to division.
There are several ways to go about solving this problem. L'Hopital rule is one way to go, however, it a little bit
overkill. It goes like this:
[tex]\lim_{n \rightarrow \infty} \left( \sin \left( 1 + \frac{\pi}{n} \right) + n \sin \left( \frac{\pi}{n} \right) \right) = \lim_{n \rightarrow \infty} \sin \left( 1 + \frac{\pi}{n} \right) + \lim_{n \rightarrow \infty} n \sin \left( \frac{\pi}{n} \right) = \sin (1) + \lim_{n \rightarrow \infty} \frac{\sin \left( \frac{\pi}{n} \right)}{\frac{1}{n}}[/tex]
Now, can you apply L'Hopital to this problem?
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Another nicer way is to use the well-know limit:
[tex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/tex]
The first few steps are just the same:
[tex]\lim_{n \rightarrow \infty} \left( \sin \left( 1 + \frac{\pi}{n} \right) + n \sin \left( \frac{\pi}{n} \right) \right) = \lim_{n \rightarrow \infty} \sin \left( 1 + \frac{\pi}{n} \right) + \lim_{n \rightarrow \infty} n \sin \left( \frac{\pi}{n} \right) = \sin (1) + \lim_{n \rightarrow \infty} \frac{\sin \left( \frac{\pi}{n} \right)}{\frac{1}{n}}[/tex]
As [tex]n \rightarrow \infty[/tex], we have [tex]\frac{1}{n} \rightarrow 0[/tex], right? Let t = 1 / n, so your limit becomes:
[tex]... = \sin 1 + \lim_{t \rightarrow 0} \frac{\sin (\pi t)}{t}[/tex], now, you'll try to rearrange your limit so that it would look the same as:
[tex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/tex]
Multiply both numerator, and denominator by pi, yields:
[tex]... = \sin 1 + \pi \lim_{t \rightarrow 0} \frac{\sin (\pi t)}{\pi t}[/tex]
You should be able to go from here, right? :)