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Converge or Diverge?
- Thread starter thatguy47
- Start date
The first one definitely converges and you are correct.
Here is what it converges to:
\(\displaystyle 11\cdot \frac{\frac{3}{7}}{1+\frac{3}{7}}=\frac{-33}{10}\)
As for the second one, yes it converges as well. You are correct.
What does it converge to?.
\(\displaystyle \frac{6(sin^{2}(1))}{8sin^{2}(1)+1}\)
Here is what it converges to:
\(\displaystyle 11\cdot \frac{\frac{3}{7}}{1+\frac{3}{7}}=\frac{-33}{10}\)
As for the second one, yes it converges as well. You are correct.
What does it converge to?.
\(\displaystyle \frac{6(sin^{2}(1))}{8sin^{2}(1)+1}\)
Hello, thatguy47!
Your work is correct!
\(\displaystyle \text{This is a geomtric series: }\;11\bigg[\left(\text{-}\tfrac{3}{7}\right) + \left(\text{-}\tfrac{3}{7}\right)^2 + \left(\text{-}\tfrac{3}{7}\right)^3 + \hdots\bigg]\)
. . \(\displaystyle \text{with first term }a = \text{-}\tfrac{3}{7}\text{ and common ratio }r = \text{-}\tfrac{3}{7}\)
\(\displaystyle \text{Not only does it converge, but we can find its sum: }\:S \:=\:11\left[\frac{\text{-}\frac{3}{7}}{1-(-\frac{3}{7})}\right] \;=\;-\frac{33}{10}\)
Edit: Too fast for me, galactus!
.
Your work is correct!
\(\displaystyle 1.\;\;\sum^{\infty}_{n=1}11\left(\text{-}\tfrac{3}{7}\right)^n\)
\(\displaystyle \text{This is a geomtric series: }\;11\bigg[\left(\text{-}\tfrac{3}{7}\right) + \left(\text{-}\tfrac{3}{7}\right)^2 + \left(\text{-}\tfrac{3}{7}\right)^3 + \hdots\bigg]\)
. . \(\displaystyle \text{with first term }a = \text{-}\tfrac{3}{7}\text{ and common ratio }r = \text{-}\tfrac{3}{7}\)
\(\displaystyle \text{Not only does it converge, but we can find its sum: }\:S \:=\:11\left[\frac{\text{-}\frac{3}{7}}{1-(-\frac{3}{7})}\right] \;=\;-\frac{33}{10}\)
Edit: Too fast for me, galactus!
.