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Math Help - Series problem involving trig. function

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    Series problem involving trig. function

    I'm having trouble with 3 series questions involving trig.

    1. \sum_{n=1}^\infty \frac{cos^2 (n)}{n^2 +1}\

    2. \sum_{n=0}^\infty \frac{1+ sin (n)}{10^n}\

    3. \sum_{n=1}^\infty \frac{arctan (n)}{n^{1.2}}\

    For the most part I know that they are convergent, but I am unable to show it. I have attempted to demonstrate it using a limit comparison test, but have so far been unsuccessful.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lllll View Post
    I'm having trouble with 3 series questions involving trig.

    1. \sum_{n=1}^\infty \frac{cos^2 (n)}{n^2 +1}\
    note that \sum_{n = 1}^{\infty} \frac {\cos^2 n}{n^2 + 1} \le \sum_{n = 1}^{\infty} \frac 1{n^2} for all n \in \mathbb{N}

    use the (direct) comparison test
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lllll View Post
    2. \sum_{n=0}^\infty \frac{1+ sin (n)}{10^n}\
    similar to the last one: \sum_{n = 0}^{\infty} \frac {1 + \sin n}{10^n} \le \sum_{n = 0}^{\infty} \frac 2{10^n}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by lllll View Post
    3. \sum_{n=1}^\infty \frac{arctan (n)}{n^{1.2}}\
    same thought: \sum_{n = 1}^{\infty} \frac {\arctan x}{n^{1.2}} \le \frac {\pi}2 \sum_{n = 1}^{\infty} \frac 1{n^{1.2}}
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    Quote Originally Posted by Jhevon View Post
    similar to the last one: \sum_{n = 0}^{\infty} \frac {1 + \sin n}{10^n} \le \sum_{n = 0}^{\infty} \frac 2{10^n}
    You should say:  \left| \frac{1+\sin n}{10^n} \right| \leq \frac{2}{10^n}.

    Because the direct comparasion test does not work otherwise.

    Or, you can write 0\leq \frac{1+\sin n}{10^n} \leq \frac{2}{10^n}. Again, because if you do not include that the test does not necessary work (and it is easy to find an example).
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