Help with power series question

[jayz]

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Hi guys, So i've been stumped by this question because of the negative 5^n part. I havent encountered what to do if there is a negative sign inside as i try to solve. What i have so far is | -5(x+3) | <1 . Is that correct or did I mess something up before?

Also for (-1^n)(x^n)/(n!) , why is the answer all convergeant? I have gotton to the step where | -x/(n+1) | >1 but I dont know how to do the next step. Thank you guys!

 

[HallsofIvy]

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Hi guys, So i've been stumped by this question because of the negative 5^n part. I havent encountered what to do if there is a negative sign inside as i try to solve. What i have so far is | -5(x+3) | <1 . Is that correct or did I mess something up before?

Also for (-1^n)(x^n)/(n!) , why is the answer all convergeant? I have gotton to the step where | -x/(n+1) | >1 but I dont know how to do the next step. Thank you guys!

That's an absolute value! You understand that \(\displaystyle \left|\frac{-x}{n+1}\right|= \frac{|x|}{n+1}\), don't you?

 

[Ishuda]

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Hi guys, So i've been stumped by this question because of the negative 5^n part. I havent encountered what to do if there is a negative sign inside as i try to solve. What i have so far is | -5(x+3) | <1 . Is that correct or did I mess something up before?

Problem (1) Convergence of
\(\displaystyle \Sigma_{1}^{\infty}\frac{(-5)^n}{\sqrt{n}} (x+3)^n\)
If we do the ratio test, we see that if |-5(x+3)|=|5(x+3)|<1 we have absolute convergence and if |5(x+3)|>1, the series diverges [which is what you have]. But what about the values \(\displaystyle \pm\) 1? First, re-write the series
\(\displaystyle \Sigma_{1}^{\infty}\frac{(-5)^n}{\sqrt{n}} (x+3)^n=\Sigma_{1}^{\infty}\frac{(-1)^n(5)^n}{\sqrt{n}} (x+3)^n=\Sigma_{1}^{\infty}\frac{(-1)^n}{\sqrt{n}} [5(x+3)]^n\)
If 5(x+3) = 1, we have
\(\displaystyle \Sigma_{1}^{\infty}\frac{(-1)^n}{\sqrt{n}} 1^n=\Sigma_{1}^{\infty}\frac{(-1)^n}{\sqrt{n}}\)
Looks like a good candidate for the alternating series test.
If 5(x+3) = -1, we have
\(\displaystyle \Sigma_{1}^{\infty}\frac{(-1)^n}{\sqrt{n}} (-1)^n=\Sigma_{1}^{\infty}\frac{(-1)^n (-1)^n}{\sqrt{n}}=\Sigma_{1}^{\infty}\frac{1}{\sqrt{n}}\)
Looks like a good candidate for comparison with the harmonic series.

Also for (-1^n)(x^n)/(n!) , why is the answer all convergeant? I have gotton to the step where | -x/(n+1) | >1 but I dont know how to do the next step. Thank you guys!

Problem (2) Convergence of
\(\displaystyle \Sigma_{1}^{\infty}\frac{(-1)^n x^n}{n!}\)
You appeared to have done the ratio test properly in the last problem, what does that give you here? Why do have
\(\displaystyle |\frac{-x}{n+1}|\space =\space |\frac{x}{n+1}|\) > 1?
Is that possible in the limit as n goes to \(\displaystyle \infty\)?

 

[jayz]

Ishuda so for #1, what would be the range of covergance? would it be (-2.8,3.2) ?

 

[Ishuda]

Ishuda so for #1, what would be the range of covergance? would it be (-2.8,3.2) ?

Almost, but you made a mistake somewhere. We know, as you have already determined, the series for Problem (1) converges absolutely for
|5(x+3)| < 1
or
-1 < 5 (x+3) < 1 ==> -16 < 5x < -14 ==> -3.2 < x < -2.8
or (-3.2, -2.8).
What you haven't determined, as far as I know, is whether the series converges for x=-3.2 or for x = -2.8. If it converges for either (both) of those then that side (those sides) becomes a closed interval- [ or ] - rather than an open interval - ( or ).