Consider\(\{a_n \}_{n=1}\) and let \(s_n = \sum _{k=1}^{n} a_k\). Suppose it is known that
\[ s_n = \frac {2n+1}{3n} \]

Then, the sequence \(\{s_n\}\) is:

increasing decreasing monotonic bounded above bounded below bounded
(Select all that apply)
We can write:
\[ s_n = \frac {2n+1}{3n} = \frac {2n}{3n}+\frac {1}{3n} = \answer {\frac {2}{3}}+\frac {1}{\answer {3n}} \]
Hence, \(s_n\) is:
increasing decreasing
Since \(\{s_n\}\) is decreasing, it must be:
bounded below bounded above
We see \(\lim _{n \to \infty } s_n = \answer {\frac {2}{3}}\).

If a sequence has a limit, then it:

must have a limit. could have a limit but does not have to have a limit. must not have a limit.

The sequence \(\{a_n \}_{n=2}\) is:

increasing decreasing monotonic bounded above bounded below bounded
(Select all that apply)
Since \(s_1=a_1\), we find: \(a_1 = \answer {1}\).

Note, that after we specify the first term, we always have a recursive formula for \(s_n\):

\[ s_n = s_{n-1}+a_n \]
(we have to require \(s_1=a_1\) in order to start this recursive formula!)

We can use the formula to find and solve for \(a_n\) for \(n \geq 2\):

\[ a_n = \answer {\frac {1}{3n}-\frac {1}{3n-3}} \]
Writing this as a simplified single rational function:
\[ a_n = \frac {\answer {-3}}{9n^2-3n} \]