# Thread: definition of convergence and divergence of a series

1. ## definition of convergence and divergence of a series

1. $\displaystyle \displaystyle \sum _{n=1}^{\infty }\left[\frac{1}{n^a}-\frac{1}{(n+1)^a}\right],\ a\in \mathbb{R}$
2. $\displaystyle \displaystyle \sum _{n=1}^{\infty }\ln \left(1+\frac{1}{n}\right).$

How would I go about finding whether these series are convergent or divergent using the definition (ie. the series i=1 to infinity of ai is convergent if the limit as n approaches infinity of the series i=1 to n of ai exists, otherwise it is divergent)

any help would be appreciated thanks

2. Originally Posted by khood
1. $\displaystyle \displaystyle \sum _{n=1}^{\infty }\left[\frac{1}{n^a}-\frac{1}{(n+1)^a}\right],\ a\in \mathbb{R}$
2. $\displaystyle \displaystyle \sum _{n=1}^{\infty }\ln \left(1+\frac{1}{n}\right).$
How would I go about finding whether these series are convergent or divergent using the definition (ie. the series i=1 to infinity of ai is convergent if the limit as n approaches infinity of the series i=1 to n of ai exists, otherwise it is divergent)

any help would be appreciated thanks
Do you mean that the sequence of partial sums doesn't converge?

For the first one write out the first couple of terms and you will see a pattern. Same for the second one once you rewrite $\displaystyle \ln\left(1+\frac{1}{n}\right)=\ln(n+1)-\ln(n)$

Both of these are known as "telescoping series"

3. sorry, I wasn't overly clear
the exact question relating to parts a. and b. is "Use the definition to determine whether the following series are convergent or divergent. When a series is convergent, find the sum."

4. Originally Posted by khood
sorry, I wasn't overly clear
the exact question relating to parts a. and b. is "Use the definition to determine whether the following series are convergent or divergent. When a series is convergent, find the sum."
Exactly what I said

Define $\displaystyle S_N=\sum_{n=1}^{N}\bigg[\frac{1}{n^{\alpha}}-\frac{1}{(n+1)^{\alpha}}\bigg]$

\displaystyle \begin{aligned}&S_1=1-\frac{1}{2^n}\\ &S_2=1-\frac{1}{3^{\alpha}}\\ &S_3=1-\frac{1}{4^{\alpha}}\\ &\vdots\\ &S_N=1-\frac{1}{\left(N+1\right)^{\alpha}}\end{aligned}

So noting that $\displaystyle \lim_{N\to\infty}S_N=\sum_{n=1}^{\infty}\bigg[\frac{1}{(n+1)^{\alpha}}-\frac{1}{n^{\alpha}}\bigg]$

We can see that $\displaystyle S_{\infty}=1$

For the second one defining $\displaystyle S_N=\sum_{n=1}^{N}\bigg[\ln(n+1)-\ln(n)\bigg]$ we see that

\displaystyle \begin{aligned}&S_1=\ln(2)\\ &S_2=\ln(3)\\ &S_3=\ln(4)\\ &\vdots\\ &S_N=\ln\left(N+1\right)\end{aligned}

So

\displaystyle \begin{aligned}\sum_{n=1}^{\infty}\bigg[\ln(n+1)-\ln(n)\bigg]&=\lim_{N\to\infty}S_N\\ &=\lim_{N\to\infty}\ln\left(N+1\right)\\ &=\infty\end{aligned}

So this series diverges