0
$\begingroup$

If the $n$th partial sum of a series $$\sum_{n=1}^\infty a_n $$ is $s_n=8-n6^{-n}$, find $a_1$, $a_n$, and $$\sum_{n=1}^\infty a_n $$

What I did:

$$a_1=8-\frac { 1 }{ 6 } =\frac { 47 }{ 6 } $$

$$\sum _{ n=1 }^{ \infty }{a_n} =\lim _{ n\rightarrow \infty }{ 8-\frac {n}{6^n}}=8$$


Now, I read in Stewart's Calculus that $a_n=s_n-s_{n-1}$, so I did tried to find it by doing:

$$\lim _{ n\rightarrow \infty }{ 8-\frac {n-1}{6^{n-1}}}=\lim _{ n\rightarrow \infty }{ 8-\frac {n-1}{6^n\cdot\frac { 1 }{ 6 }}}=8$$

Then $s_n-s_{n-1}=0$? However, this doesn't seem to be correct. What am I doing wrong?

I have tried my textbook, Khan Academy, and even a few questions on this site such as this one, but I still do not understand what needs to be done. Any help/guidance would be appreciated.

$\endgroup$
  • $\begingroup$ Why the down-vote? I followed the guidelines of this website in posting this question. I posted a legitimate question. I stated what I already tried, what I don't understand, and what resources I have utilized to attempt to figure it out on my own. $\endgroup$ – Cherry_Developer Oct 16 '16 at 23:37
  • 1
    $\begingroup$ I have the impression that there are some users who downvote questions that in their opinion shouldn't have needed to be asked, or that contain errors that they think are so obviously wrong that they shouldn't have been made. It is, as you say, a perfectly good question, and I've upvoted it. $\endgroup$ – Brian M. Scott Oct 16 '16 at 23:47
1
$\begingroup$

The partial sum is $$ s_n=\sum_{k=1}^na_k\tag{1} $$ From $(1)$, we get that $a_1=s_1$ and for $n\gt1$, $$ a_n=s_n-s_{n-1}\tag{2} $$ If we know that $s_n=8-n6^{-n}$, then $a_1=\frac{47}6$ and for $n\ge2$, $$ \begin{align} a_n &=(n-1)6^{1-n}-n6^{-n}\\ &=(5n-6)\,6^{-n}\tag{3} \end{align} $$ Furthermore, $$ \begin{align} \sum_{k=1}^\infty a_k &=\lim_{n\to\infty}\sum_{k=1}^n a_k\\ &=\lim_{n\to\infty}s_n\\[3pt] &=\lim_{n\to\infty}\left(8-n6^{-n}\right)\\[3pt] &=8\tag{4} \end{align} $$

$\endgroup$
  • $\begingroup$ Thank you! For some reason, it completely slipped my mind that $s_n - s_{n-1}$ just meant to simply subtract without involving take their limits first. $\endgroup$ – Cherry_Developer Oct 16 '16 at 23:31
1
$\begingroup$

You took $\lim_{n\rightarrow\infty}s_{n-1}$ and subtracted that from $\lim_{n\rightarrow\infty}s_n$. What you should do instead is just take $s_n-s_{n-1}$. That is, $a_n=\left(8-\frac{n}{6^n}\right)-\left(8-\frac{n-1}{6^{n-1}}\right)=\frac{5n-6}{6^n}$.

$\endgroup$
0
$\begingroup$

It is not an equality of $\;s_n-s_{n-1}=0\;$ but of their limit! Why not correct? The series converges$\;\iff \lim\limits_{n\to\infty}s_n=S\;$ is finite (and exists, of course), and thus

$$\lim_{n\to\infty} a_n=\lim_{n\to\infty}(s_n-s_{n-1})=S-S=0$$

and it is a well known necessary, though not sufficient, condition for a series to converge that its general sequence's limit is zero.

$\endgroup$
  • $\begingroup$ So, am I correct in saying that $a_n=0$? $\endgroup$ – Cherry_Developer Oct 16 '16 at 23:12
  • $\begingroup$ @Cherry_Developer No. What is true is that $\;\lim\limits_{n\to\infty} a_n=0\;$ . It's not the same ... $\endgroup$ – DonAntonio Oct 16 '16 at 23:13
  • $\begingroup$ I apoligize, but now I am very confused. How would I use what I am given and have already figured out to find $a_n$? $\endgroup$ – Cherry_Developer Oct 16 '16 at 23:17
  • $\begingroup$ @Cherry_Developer You already found $\;a_1\;$ , you can easily find out $\;a_n\;$ doing $\;s_n-s_{n-1}\;$ and you've already found out what the sum of the (comvergent, of course) series is. What else do you want? $\endgroup$ – DonAntonio Oct 16 '16 at 23:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.