For what value(s) of r is the series the summation from n=1 to infinity of 1/n^r+1 convergent?
a) r > -2
b) r > -1
c) r ≥ -2
d) All values of r
do you mean 1/n^(r+1)
well d is just plain out
1/n^(-100 + 1 ) gets big
if r = -2
looks like
1/ n^(-2 + 1) = n hardly
if r = -1
1/ (n^0) = 1 that works and so does any bigger r
Sorry it was my fault. The question is: For what value(s) of r is the series the summation from n=1 to infinity of 1/(n^r+3) convergent?
To determine for what values of r the series \(\sum_{n=1}^{\infty} \frac{1}{n^{r+1}}\) is convergent, we can use the p-series test.
The p-series test states that for any series of the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\), where p is a positive real number, the series is convergent if p > 1 and divergent if p ≤ 1.
In our case, we have the series \(\sum_{n=1}^{\infty} \frac{1}{n^{r+1}}\). By comparing it to the general form of a p-series, we can see that p = r+1.
According to the p-series test, the series will be convergent if p > 1. Thus, we need to determine the values of r for which r+1 > 1.
Simplifying the inequality, we have r > 0. Subtracting 1 from both sides, we get r > -1.
Therefore, the correct answer is b) r > -1, which means that the series \(\sum_{n=1}^{\infty} \frac{1}{n^{r+1}}\) is convergent for all values of r greater than -1.