Find a formula for the partial sums of the series $$\sum _{n=1}^\infty \left [ e^{\frac {1}{n}} - e^{\frac {1}{n+1}}\right ]$$ and then compute the sum of the series.

• We can start by computing the first few partial sums: $$\begin {aligned} S_1 & = (e - e^{1/2}) \\ S_2 & = (e^{1} - e^{1/2}) + (e^{1/2} - e^{1/3}) = e - e^{1/3} \\ S_3 & = (e^{1} - e^{1/2}) + (e^{1/2} - e^{1/3}) + (e^{1/3} - e^{1/4}) = e - e^{1/4} \\ S_4 & = (e^{1} - e^{1/2}) + (e^{1/2} - e^{1/3}) + (e^{1/3} - e^{1/4}) + (e^{1/4} - e^{1/5}) = e - e^{1/5} \end {aligned}$$ We observe a general pattern that is consistent with a telescoping series: $$S_n = e - \answer {e^{1/(n+1)}}$$
• Now we let $n \rightarrow \infty$: $$\lim _{n \rightarrow \infty } S_n = \lim _{n \rightarrow \infty } \left ( \answer {e - e^{1/(n+1)}} \right ) = e - e^{\lim _{n \rightarrow \infty } \frac {1}{n+1}} = e - e^{\answer {0}} = e - 1.$$ Therefore $$\sum _{n=1}^\infty \left [ e^{\frac {1}{n}} - e^{\frac {1}{n+1}}\right ] = \answer {e-1}.$$

Show that each series below is a geometric series; determine which ones are convergent, and for those that are convergent, find their sum. $$\sum _{n=0}^\infty \frac {3^n}{2} \qquad \sum _{n=0}^\infty \frac {2}{3^n} \qquad \sum _{n=0}^\infty 2^{-n-2} \qquad \sum _{n=0}^\infty \frac {(-1)^{n+1} 3^n}{4^n}$$

• First we rewrite each geometric series in standard form: i.e., write each term in the form $a r^n$ where $a$ and $r$ are fixed quantities independent of $n$: $$\sum _{n=0}^\infty \frac {3^n}{2} = \sum _{n=0}^\infty \answer {\frac {1}{2}} \left [ \answer {3} \right ]^n$$ $$\sum _{n=0}^\infty \frac {2}{3^n} = \sum _{n=0}^\infty \answer {2} \left [ \answer {\frac {1}{3}} \right ]^n$$ $$\sum _{n=0}^\infty 2^{-n-2} = \sum _{n=0}^\infty \answer {\frac {1}{4}} \left [ \answer {\frac {1}{2}} \right ]^n$$ $$\sum _{n=0}^\infty \frac {(-1)^{n+1} 3^n}{4^n} = \sum _{n=0}^\infty \answer {-1} \left [ \answer {\frac {-3}{4}} \right ]^n$$
• A geometric series is convergent exactly when $|r| < \answer {1}$. This means that the following series are convergent (select all that apply):
  $\displaystyle \sum _{n=0}^\infty \frac {3^n}{2}$ $\displaystyle \sum _{n=0}^\infty \frac {2}{3^n}$ $\displaystyle \sum _{n=0}^\infty 2^{-n-2}$ $\displaystyle \sum _{n=0}^\infty \frac {(-1)^{n+1} 3^n}{4^n}$
• This means that the given series have the sums (write N/A if the series diverges, otherwise give the sum) $$\sum _{n=0}^\infty \frac {3^n}{2} = \answer {N/A} \qquad \sum _{n=0}^\infty \frac {2}{3^n} = \answer {3} \qquad \sum _{n=0}^\infty 2^{-n-2} = \answer {\frac {1}{2}} \qquad \sum _{n=0}^\infty \frac {(-1)^{n+1} 3^n}{4^n} = \answer {-\frac {4}{7}}.$$

Reindex each series below to begin at the specified starting value:

Series 1:

$$\sum _{n=0}^\infty \frac {1}{n+1} = \sum _{n=1}^\infty \answer {\frac {1}{n}}$$

Series 2:

$$\sum _{n=2}^\infty e^{-2n-1} = \sum _{n=0}^\infty e^{\answer {-2n-5}}$$

Series 3:

$$\sum _{n=3}^\infty (-1)^n \cos \frac {n^2 + n}{3} = \sum _{n=2}^\infty \answer {(-1)^{n+1}} \cos \answer {\frac {(n+1)^2+(n+1)}{3}}$$

Suppose the series $\displaystyle \sum _{n=1}^\infty a_n$ and $\displaystyle \sum _{n=1}^\infty b_n$ both converge and the series $\displaystyle \sum _{n=1}^\infty c_n$ and $\displaystyle \sum _{n=1}^\infty d_n$ both diverge. Then

• The series $\displaystyle \sum _{n=1}^\infty (a_n + b_n)$ convergesdivergesmay converge or diverge .
• The series $\displaystyle \sum _{n=1}^\infty (a_n + c_n)$ convergesdivergesmay converge or diverge .
• The series $\displaystyle \sum _{n=1}^\infty (c_n + d_n)$ convergesdivergesmay converge or diverge .