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Determine whether this is convergent or not: $$\sum _{n=1}^{\infty }\:\frac{100\cdot \:101\cdot \:...\cdot \:\left(100+n\right)}{1\cdot 3\cdot ...\cdot \left(2n-1\right)}$$

My problem is that I didn't really work with such series before, where I have $...$ inside the sum. I'm thinking I can just use the ratio test and it would tell me that my series is convergent. Is this the correct approach?

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  • $\begingroup$ Do you mean $ \sum _{n=1}^{\infty }\:\frac{100\cdot\color{red}{101}\cdots\left(100+n\right)}{1\cdot 3\cdot ...\cdot \left(2n-1\right)} $? Also, for $n = 1$, is the intended fraction $\frac{100\cdot 101}{1}$? Because that's what it looks like (in contrast to, let's say, $\frac{100}{1}$ or $\frac{100\cdot 101}{1\cdot 3}$). $\endgroup$ – Arthur Nov 11 '15 at 10:07
  • $\begingroup$ Yes! Sorry for the error, didn't notice. $\endgroup$ – MikhaelM Nov 11 '15 at 10:09
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Hint

Do not worry too much about the complexity of the terms. You need to work $$S=\sum_{i=1}^\infty a_n$$

Consider $$a_n=\frac{100\times 101\times \cdots\times \left(100+n\right)}{1\times 3\times \cdots\times \left(2n-1\right)}$$ $$a_{n+1}=\frac{100\times 101\times \cdots\times \left(100+n\right)\times(101+n)}{1\times 3\times \cdots\times \left(2n-1\right)\times(2n+1)}$$ So, the ratio of two consecutive terms is just $$\frac{a_{n+1}}{a_{n}}=\frac{n+101}{2 n+1}$$

I am sure that you can take from here.

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  • $\begingroup$ Yes! This is how I solved it, but I didn't know if this was the correct approach. Thank you. $\endgroup$ – MikhaelM Nov 11 '15 at 10:30
  • $\begingroup$ You are welcome ! It is simple; just use the definition and the ususal tests. $\endgroup$ – Claude Leibovici Nov 11 '15 at 10:33
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To prove it, don't be impressed with the $\dots$.

Take $N = 100$,

Split summation up to $N$ and above $N$:

$\sum\limits_{i=1}^{\infty} X_i = \sum\limits_{i=1}^{100} X_i + \sum\limits_{i=101}^{\infty} X_i$

The first part is convergent as is any finite summation.

The second part can be bounded by a constant (even if it is $200^{100}$) times a convergent sequence $1/2^n$.

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  • $\begingroup$ I don't understand what you mean with 'split summation up to N and above N' :( $\endgroup$ – MikhaelM Nov 11 '15 at 10:13

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