- #1
taekwondo22
- 10
- 0
1) Determine whether the series converges or diverges: summation from n=1 to ∞ of (square root of (n+1) - square root of (n-1)) / n. clearly state which test you are using.
2) Determine whether the series converges or diverges: summation from n=1 to ∞ of (1*3*5*... (2n-1)) / (2*5*8*... (3n-1)). clearly state which test you are using.
For question #1, I tried multiplying the top and bottom by square root of (n+1) + square root of (n-1). On the top, the answer simplifies to 2 and on the bottom it simplifies to n multiplied by (square root of (n+1) + square root of (n-1)). I am thinking to divide the top and bottom by n so the limit as n approaches infinity is equal to 0. But by the nth term test for divergence, if the limit is equal to 0, then the series may converge or diverge. This is where I am stuck and can't think of anything else.
For question #2, I am having trouble simplifying the problem. It can't be just (2n-1) / (3n-1) because that would change the whole series.
2) Determine whether the series converges or diverges: summation from n=1 to ∞ of (1*3*5*... (2n-1)) / (2*5*8*... (3n-1)). clearly state which test you are using.
For question #1, I tried multiplying the top and bottom by square root of (n+1) + square root of (n-1). On the top, the answer simplifies to 2 and on the bottom it simplifies to n multiplied by (square root of (n+1) + square root of (n-1)). I am thinking to divide the top and bottom by n so the limit as n approaches infinity is equal to 0. But by the nth term test for divergence, if the limit is equal to 0, then the series may converge or diverge. This is where I am stuck and can't think of anything else.
For question #2, I am having trouble simplifying the problem. It can't be just (2n-1) / (3n-1) because that would change the whole series.
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