What's the defining characteristic of a geometric series? A series \(\displaystyle \sum_{n=0}^{\infty} a_n\) is geometric if ______________

If the ratio of successive terms of a positive series eventually approaches a constant \(L<1\), then the series can be bounded above by a convergent geometric series with ratio \(0<r<1\).

Theorem: Let \(\lbrace a_n \rbrace\) be a positive sequence where \(\displaystyle\lim_{n\to \infty} \frac{a_{n+1}}{a_n} = L\).

1. If \(L<1\), then the series \(\displaystyle\sum_{n=1}^{\infty} a_n\) converges.

2. If \(L>1\), then the series \(\displaystyle\sum_{n=1}^{\infty} a_n\) diverges.

3. If \(L=1\), then the ratio test is inconclusive, so try another method.

Informal summary: If a series is eventually positive and eventually like a convergent geometric series, it converges.

Choose a fixed \(r\) so that \(L < r < 1\). Then \(\displaystyle \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L\) implies there is some \(k\) so that \(a_{n+1} < r a_n\) for all \(n > k\). Thus, \[\begin{array}{l} a_{k+1} \leq r a_k \\ a_{k+2} \leq r a_{k+1} \leq r^2 a_k \\ a_{k+3} \leq r a_{k+2} \leq r^2 a_{k+1} \leq r^3 a_k \end{array}\] and hence the tail \[a_k + a_{k+1} + a_{k+2} + a_{k+3} + \cdots\] is bounded above by a convergent geometric series \[a_k + r a_k + r^2 a_k + r^3 a_k + \cdots.\]

Determine whether these series converge. Don't forget to check validity conditions!

1. \(\displaystyle \sum_{n=1}^{\infty} \frac{n! n^2}{(2n)!}\)

2. \(\displaystyle \sum_{n=1}^{\infty} \frac{5^n-3n}{4^n}\)

In a geometric series, \(a_n = r^n = r \cdot r \cdot r \cdots r\) (with \(n\) factors), and thus \(\sqrt[n]{a_n} = (r^n)^{1/n} = r\).

In an arbitrary series with all positive terms, split each term, \(a_n\), of the series into \(n\) equal size factors. If each factor \(\sqrt[n]{a_n}\) is less than a fixed positive ratio \(r<1\), then the series is bounded above by a convergent geometric series.

Theorem: Let \(\lbrace a_n \rbrace\) be a positive sequence where \(\displaystyle\lim_{n\to \infty} \sqrt[n]{a_n} = L\).

1. If \(L<1\), then the series \(\displaystyle\sum_{n=1}^{\infty} a_n\) converges.

2. If \(L>1\), then the series \(\displaystyle\sum_{n=1}^{\infty} a_n\) diverges.

3. If \(L=1\), then the ratio test is inconclusive, so try another method.

Informal summary: If a series is eventually positive and eventually like a convergent geometric series, it converges.

Choose a fixed \(r\) so that \(L < r < 1\). Then \(\displaystyle \lim_{n \to \infty} \sqrt[n]{a_n} = L\) implies there is some \(k\) so that \(\sqrt[n]{a_n} < r\) for all \(n > k\). Thus, \(a_n < r^n\) for all \(n > k\) and hence the tail \[a_k + a_{k+1} + a_{k+2} + a_{k+3} + \cdots\] is bounded above by a convergent geometric series \[r^k + r^{k+1} + r^{k+2} + r^{k+3} + \cdots.\]

\(\displaystyle \lim_{n \to \infty} \sqrt[n]{C} = 1\) for any constant \(C\)

\(\displaystyle \lim_{n \to \infty} \sqrt[n]{n^p} = 1\) for any \(p>0\)

\(\displaystyle \lim_{n \to \infty} \sqrt[n]{b^n} = b\) for any constant \(b\)

\(\displaystyle \lim_{n \to \infty} \sqrt[n]{n!} = \infty\)

\(\displaystyle \lim_{n \to \infty} \sqrt[n]{n^n} = \infty\)

Determine whether these series converge. Don't forget to check validity conditions!

1. \(\displaystyle \sum_{n=1}^{\infty} \left( \frac{n+2}{2n+1} \right)^n\)

2. \(\displaystyle \sum_{n=0}^{\infty} \frac{5^n (3n^2+1)}{2^n n!}\)

3. \(\displaystyle \sum_{n=2}^{\infty} \frac{n^2}{(\ln n)^n }\)