help with a series question

[soccerball3211]

Given the power series, the summation from n=0 to infinity of (a sub n times x^n)
where a sub 0 = (n/n+1) times (a sub n -1) for n is greater than or equal to 1

a.) find the first 5 terms of the series and the general term
b.) for what values of x does the series converge
c.) if g(x)=the summation from n=0 to infinity of (a sub n times x^n), find g'(1)

 

[pka]

Given the power series, the summation from n=0 to infinity of (a sub n times x^n)
where a sub 0 = (n/n+1) times (a sub n -1) for n is greater than or equal to 1

Please correct the definition of the terms.
As written it cannot be.
I think mean \(\displaystyle n \ge 1\quad \Rightarrow \quad a_n = \frac{n}{{n + 1}}a_{n - 1}\).
But then you need to tell us what \(\displaystyle a_0\) equals.

 

[soccerball3211]

my mistake, a sub 0 =1
and a sub n = (n/n+1) times (a sub n -1)

sorry about the mistake

 

[pka]

I will get you started.
\(\displaystyle \L
\begin{array}{l}
a_0 = 1 \\
a_1 = \frac{1}{{1 + 1}}a_0 = \frac{1}{2} \\
a_2 = \frac{2}{{2 + 1}}a_1 \\
\end{array}\).

Now you show how much you can do.

 

[soccerball3211]

for a sub 2 I got 1/6
for a sub 3 I got 1/24
for a sub 4 I got 1/120
for a sub 5 I got 1/720

for the general term I got (a sub n-1) times x^(n-1) + (a sub n) times x^n

i can't figure out the radius of convergence

 

[pka]

TRY AGAIN!
\(\displaystyle \L
\begin{array}{l}
a_0 = 1 \\
a_1 = \frac{1}{{1 + 1}}a_0 = \frac{1}{2} \\
a_2 = \frac{2}{{2 + 1}}a_1 = \frac{2}{{2 + 1}}\left( {\frac{1}{2}} \right) = \frac{1}{3} \\
a_3 = \frac{3}{{3 + 1}}a_2 = \frac{3}{4}\left( {\frac{1}{3}} \right) = \frac{1}{4} \\
\end{array}\)

 

[soccerball3211]

is the general term n/(n+1)?

 

[pka]

is the general term n/(n+1)?

Before I tell you, you must correctly tell me:
\(\displaystyle \L a_4 = ?\\
a_5 = ?\)

 

[soccerball3211]

a sub 4 = 1/5
a sub 5 = 1/6

 

[pka]

Good! Now look at the subscript and look at the term.
Would bet me that \(\displaystyle a_{11} = \frac{1}{12}\)?

What is \(\displaystyle a_n\) ?

 

[soccerball3211]

1/(n+1)

 

[pka]

YES! Now we the power series:
\(\displaystyle \L
\sum\limits_{n = 0}^\infty {a_n x^n } = \sum\limits_{n = 0}^\infty {\frac{{x^n }}{{n + 1}}}\)
to work with.

 

[soccerball3211]

ok pka,
I applied the ratio test and got abs(x)
therefore abs(x)<1

so when x=+1 the series becomes 1+(1/2)+(1/3)+... diverges

and when x=-1 the series becomes 1-(1/2)+(1/3)-(1/4)+... converges

interval of convergence = -1which is less than or equal to x witch is less than 1

PKA did i do this right

 

[pka]

I applied the ratio test and got abs(x)
therefore abs(x)<1

so when x=+1 the series becomes 1+(1/2)+(1/3)+... diverges

and when x=-1 the series becomes 1-(1/2)+(1/3)-(1/4)+... converges

interval of convergence = -1which is less than or equal to x which is less than

What a good job! Full marks for you!

 

[soccerball3211]

can you help get me started on part c?

 

[pka]

\(\displaystyle \L
g(x) = \sum\limits_{n = 0}^\infty {\frac{{x^n }}{{n + 1}}} \quad \Rightarrow \quad g'(x) = \sum\limits_{n = 0}^\infty {\frac{{nx^{n - 1} }}{{n + 1}}}\)

 

[soccerball3211]

i got g'(1)=the summation of n=0 to infinity of (n/(n+1))