Use the limit comparison test to determine the convergence or divergence of the series.

Problem posted below.

Below that is the solution that I can't seem to wrap my head around. I don't really understand why (1/n) was used to compare with it. I also don't understand how they solved the limit.

Question

Use the limit comparison test to determine the convergence or divergence of the series. 

Problem posted below. 

Below that is the solution that I can't seem to wrap my head around. I don't really understand why (1/n) was used to compare with it. I also don't understand how they solved the limit.

anonymous · Answer

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anonymous · Answer

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ganeshie8 · Answer

what does limit comparison test say ?

anonymous · Answer

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ganeshie8 · Answer

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$\sum a_n = \sum \frac{1}{n}$ and  $\sum b_n = \sum\sin(\frac{1}{n})$

ganeshie8 · Answer

the first one is harmonic series and you must be knowing the harmonic series diverges, yes ?

anonymous · Answer

Sorry, I had to go pick up my sister from the library!! Then she wanted to go get some milktea with friends.... ;'((((

anonymous · Answer

$$\sum_{}^{}a _{n} = sin(1/n)$$ 
$$\sum_{}^{}b _{n} = (1/n)$$ 
 After following the theorem with several other problems that I do understand, this is all I know due to repetition. 
Yes, I do know that the harmonic series with 1/n is divergent but I do not understand why they used it to compare to sin(1/n).

ganeshie8 · Answer

I think that idea came from the well known limit $\lim\limits_{x\to \infty}\dfrac{\sin x}{x} = 1$

Read limit comparison test again. Since the limit is finite and the harmonic series diverges, the given series also diverges.

ganeshie8 · Answer

focus on this part http://gyazo.com/4bc76332d2a9748b0d2c1727b136b8d0

anonymous · Answer

wait is there a way to calculate where (1/n) came from? otherwise, i dont understand the logic behind it

ganeshie8 · Answer

whats the general term of given series ?

anonymous · Answer

is it sin(1/n)?

ganeshie8 · Answer

yes the idea of 1/n comes from the stuff inside sin

anonymous · Answer

ohhh, so from trig, just use the stuff inside? i thought it had to do something with highest powers of n haha :P

anonymous · Answer

okay, Thank you!!! Sorry for my numb nutness

ganeshie8 · Answer

we just want some other series to compare against
and SUM 1/n is our first guess as it is inside sin.. and also the knowledge about limit lim sinx/x.. 

you may use ANY diverging series that allows you to find the limit easily...

anonymous · Answer

oooo i see!! wow, that makes a lot more sense. right... we can use any to test if it would work and yes, i forgot about sinx/x hehe... :( as you can see, i live up to my name! xD i wish i could get things easily. ;o;