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or compare the given series with some series which is known to be divergent, as
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(harmonic series)
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(p series)
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Illustrative Example 1. Test the series
![{\displaystyle 1+{\frac {1}{1!}}+{\frac {1}{2!}}+{\frac {1}{3!}}+{\frac {1}{4!}}+\cdots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1802f626bfb2dc0707bfbafc28cf5b3897aa80f)
Solution. Here
![{\displaystyle u_{n}={\frac {1}{\left(n-1\right)!}},\;u_{n+1}={\frac {1}{n!}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b69edef6e85239a3fad0970624e498c2c9eadb18)
![{\displaystyle \therefore \lim _{n\to \infty }\left({\frac {u_{n+1}}{u_{n}}}\right)=lim_{n\to \infty }\left({\frac {\frac {1}{n}}{\frac {1}{\left(n-1\right)!}}}\right)=\lim _{n\to \infty }\left({\frac {\left(n-1\right)!}{n!}}\right)=\lim _{n\to \infty }\left({\frac {1}{n}}\right)=0\left(=p\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f1887ae215e72d4331ae2cb1111f12ed55f4f6c)
and by
I, §141, the series is convergent.
Illustrative Example 2. Test the series
![{\displaystyle {\frac {1!}{10}}+{\frac {2!}{10^{2}}}+{\frac {3!}{10^{3}}}+\cdots }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8e322976f0ceca9a2bb72a0f3f53ab7cc2b1f31)
Solution. Here
![{\displaystyle u_{n}={\frac {n!}{10^{n}}},\;u_{n+1}={\frac {\left(n+1\right)!}{10^{n+1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6505e835e202e9c1f1cfb3cdb8dfb3f6f3806e26)
![{\displaystyle \therefore \lim _{n\to \infty }\left({\frac {u_{n+1}}{u_{n}}}\right)=\lim _{n\to \infty }\left({\frac {\left(n+1\right)!}{10^{n+1}}}\times {\frac {10^{n}}{n!}}\right)=\lim _{n\to \infty }\left({\frac {n+1}{10}}\right)=\infty \left(=p\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/23839db2e3b28bbb384dbae99af3191b2aba81f4)
and by II,§141 the series is divergent.
Illustrative Example 3. Test the series
(C)
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Solution. Here
![{\displaystyle u_{n}={\frac {1}{\left(2n-1\right)2n}},\;u_{n+1}{\frac {1}{\left(2n+1\right)\left(2n+2\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4e4c9239b685516b0d24a5bd6221360c665d5e7)
![{\displaystyle \therefore \lim _{n\to \infty }\left({\frac {u_{n+1}}{u_{n}}}\right)=\lim _{n\to \infty }\left[{\frac {\left(2n-1\right)2n}{\left(2n+1\right)\left(2n+2\right)}}\right]={\frac {\infty }{\infty }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22d58ae7751a0b4bcfce9335826b8de44efc3f4e)
This being an indeterminate form, we evaluate it, using the rule in §112.
Differentiating,
![{\displaystyle \lim _{n\to \infty }\left({\frac {8n-2}{8n-6}}\right)={\frac {\infty }{\infty }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5d4559f1c903b5d35b9b0dfd844221bfa6768f4)
Differentiating again,
![{\displaystyle \lim _{n\to \infty }\left({\frac {8}{8}}\right)=1\left(=p\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fca5ef4e297cbea719fccf94c225df456475e15d)
This gives no test (III, §141). But if we compare series (C) with (G), §138,making
, namely,
(D)
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we see that (C) must be convergent, since its terms are less than the corresponding terms of (D), which was proved convergent.