Determine the pattern for the series $\frac{1}{4} + \frac{1}{6} + \frac{1}{9} + \cdots$, then write out $S_2$, $S_4$, and $S_6$.
This is a geometric series, so we need to figure out the ratio.
Each term is $\frac{2}{3}$ times the previous one.
Thus, the generic term in the series is $\frac{1}{4} \left(\frac{2}{3}\right)^n$ starting at $n=0$.
The series can be written $\sum_{n=0}^\infty \frac{1}{4} \left(\frac{2}{3}\right)^n$, and the partial sums are $S_2 = \frac{5}{12}$, $S_4 = \frac{65}{108}$, $S_6 = \frac{665}{972}$.
Determine if $\displaystyle \sum_{n=3}^\infty \frac{3}{n(n+1)}$ converges, and if so, evaluate the sum.
This is in the same form as the telescoping series that we discussed previously. Try to write it in a form where terms will cancel.
This series can be rewritten as $\displaystyle \sum_{n=3}^\infty \frac{1}{n} - \frac{1}{n+1}$.
Write out the first few terms of this series. What can you get for a general formula for $S_N$?
$S_N = 1 - \frac{3}{N+1}$
This series converges, and the value is $1$.
Determine if $\displaystyle \sum_{n=2}^\infty \frac{1}{3} 2^n$ converges, and if so, evaluate the sum.
What type of series is this?
This is a geometric series. What is the ratio between consecutive terms?
The common ratio here is the part raised to the $n$ power, or $2$. What does that mean for the series?
This series diverges.
Determine if $\displaystyle \sum_{n=1}^\infty 2 \left(\frac{1}{3}\right)^n$ converges, and if so, evaluate the sum.
What type of series is this?
This is a geometric series. What is the ratio between consecutive terms?
The common ratio here is the part raised to the $n$ power, or $\frac{1}{3}$. What does that mean for the series?
So, we know the series converges, and is a geometric series. The formula for the sum is then $\displaystyle \frac{c}{1-r}$ where $c$ is the first term of the series. (Or the $r^0$ term if the sum starts at 0)
This series converges to $\displaystyle \frac{2/3}{1 - \frac{1}{3}} = 1$.
Determine if $\displaystyle \sum_{n=1}^\infty \frac{2n^2 + 3}{n(n+1)}$ converges, and if so, evaluate the sum.
This doesn't look like any of the types of series that we know how to evaluate. What is the only other trick we have so far?
We could try the nth term divergence test. What happens to the terms as $n \rightarrow \infty$?
If we take the limit of the terms, we get 2. What does this mean?
This series diverges by the nth term divergence test.
Determine if $\displaystyle \sum_{n=3}^\infty 4 \left(\frac{1}{5}\right)^n$ converges, and if so, evaluate the sum.
This is a geometric series. What is the ratio here?
The common ratio is $1/5$, so the series will converge. How do we evaluate the sum?
The formula for $\displaystyle \sum_{n=0}^\infty cr^n$ is $\displaystyle \frac{c}{1-r}$. That is not the exact series we have here. We need to shift the index to zero.
After shifting, we should get that the series is equivalent to $\displaystyle \sum_{m=0}^\infty \frac{4}{125} \left(\frac{1}{5}\right)^m$ by setting $m = n-3$. We can then use this for $\frac{c}{1-r}$. Notice how the $c$ for this new series is the first term of the original one. This is not a coincidence.
The series converges to $\frac{4/125}{1 - 1/5} = \frac{4/125}{4/5} = \frac{1}{25}$.
Determine if $\displaystyle \sum_{n=4}^\infty \frac{1}{n(n-2)}$ converges, and if so, evaluate the sum.
Determine if $\displaystyle \sum_{n=2}^\infty \frac{1}{2} \left(\frac{3}{5}\right)^n$ converges, and if so, evaluate the sum.