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Math Help - summation of a series

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    summation of a series

    I'm trying to compute \sum_{i=0}^{\infty}\frac{1}{(n+1)\pi}. I know the answer is \infty, but not sure how to prove this. can anyone help? thank you.
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    Quote Originally Posted by dori1123 View Post
    I'm trying to compute \sum_{i=0}^{\infty}\frac{1}{(n+1)\pi}. I know the answer is \infty, but not sure how to prove this. can anyone help? thank you.

    \sum_{i=0}^{\infty}\frac{1}{(n+1)\pi}=\frac{1}{\pi  }\sum_{i=1}^{\infty}\frac{1}{n} , and I hope you already know the harmonic series diverges so...

    Tonio
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    MHF Contributor Drexel28's Avatar
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    Personally, I find that Cauchy's condensation test is really nice here. Since \left\{\frac{1}{n}\right\}_{n=1}^{\infty} is positive and non-increasing it is true that the series \sum_{n=1}^{\infty}\frac{1}{n} shares convergence/divergence with \sum_{n=1}^{\infty}2^n\frac{1}{2^n}
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    Quote Originally Posted by RockHard View Post
    Also to further tonio's example because I need to learn this as well, you can the limit of his solution which can be easily done by taking the derivative

    \sum_{i=1}^{\infty}\frac{1}{n}<br />

    but first we know by one theorem you can write the sequence, lets called it a_n, as a f(x)
    \lim_{x\to\infty}\frac{1}{x}<br />

    take the derivative of this function which is simply

    \ln(x)

    then we have \lim_{x\to\infty}\ln(x)<br />

    Which we should know but remembering the graph of this function or plotting out some terms on a graph as X increases for values greater than 0 reaches no finite term, aka positive infinity

    Apart from the fact that what you wrote doesn't seem to be true in the general case (though a simmilar asymptotic behaviour here happens, if I understood correctly what you probably meant to say), you have some serious mistakes here: the derivative of \frac{1}{x} is -\frac{1}{x^2}.\,\,\ln x is its antiderivative or primitive function.

    Perhaps you tried to come up with a rather sui generis "proof" in the spirit of the integral test, but nevertheless it isn't quite so, and some conditions must be fulfilled in the general case.

    Tonio
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