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If $$s_n={5-{n \over 3^n}}$$ is the partial sum of the series of $\sum_{k=1}^∞ a_k$,

can we find, a-) the general term of the series b-) the sum of the series?

2 Answers 2

1

Hint:

$$a_n=s_n-s_{n-1}$$

$$\sum_{k=1}^\infty a_k=\lim_{k\to\infty}s_k$$

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    So the sum is 5. What about the general term? – 2017-02-21
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    Have you tried my first hint? It's pretty direct. – 2017-02-21
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    $$s_n=5- n/3^n $$ $$s_{(n-1)}=5-{(n-1)}/3^{(n-1)}$$ – 2017-02-21
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    You actually want $s_{n\color{red}{-1}}$. – 2017-02-21
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    Sorry. I corrected it. – 2017-02-21
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    Subtract them and simplify. – 2017-02-21
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    I found $\sum_{n=1}^∞ a_n = \frac{(2n+3)}{3^n}$ – 2017-02-21
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    That's clearly not right. Surely the right side can't have $n$ in it. – 2017-02-21
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    Sorry. It was supposed to be $$a_n= (2n-3)/3^n$$ – 2017-02-21
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    Well, there shouldn't be a summation, and then it's all good and well. – 2017-02-21
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    :-) No problem. – 2017-02-21
1

$$s_{n+1}=\dfrac{10}{3}+\frac13s_n-\frac{1}{3^{n+1}}$$ with $\lim_\infty s_n=\ell$ then $\ell=\dfrac{10}{3}+\dfrac13\ell$ so $\ell=5$.

$a_n=s_{n}-s_{n-1}=\dfrac{2n-3}{3^n}$.

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    Thanks. What about the general term? – 2017-02-21
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    Hm, a strange way of doing. If I may point out, you quietly assume the limit exists. – 2017-02-21
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    @MyGlasses lol, well, see if you can figure out why my hints work. – 2017-02-21
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    Well, it looks good now. – 2017-02-21