14
$\begingroup$

I'm a bit confused as to this problem:

Consider the infinite series:

$$\left(1 - \frac 12\right) + \left(\frac 12 - \frac 13\right) + \left(\frac 13 - \frac 14\right) \cdots$$

a) Find the sum $S_n$ of the first $n$ terms.

b) Find the sum of this infinite series.

I can't get past part a) - or rather I should say I'm not sure how to do it anymore.

Because the problem asks for a sum of the infinite series, I'm assuming the series must be Geometric, so I tried to find the common ratio based on the formula

$$r = S_n / S_{n-1} = \frac{\frac 12 - \frac 13}{1 - \frac 12} = \frac 13$$

Which is fine, but when I check the ratio of the 3rd and the 2nd terms:

$$r = \frac{\frac 13 - \frac 14}{\frac 12 - \frac 13} =\frac 12$$

So the ratio isn't constant... I tried finding a common difference instead, but the difference between 2 consecutive terms wasn't constant either.

I feel like I must be doing something wrong or otherwise missing something, because looking at the problem, I notice that the terms given have a pattern:

$$\left(1 - \frac 12\right) + \left(\frac 12 - \frac 13\right) + \left(\frac 13 - \frac 14\right) + \cdots + \left(\frac 1n - \frac{1}{n+1}\right)$$

Which is reminiscent of the textbook's proof of the equation that yields the sum of the first $n$ terms of a geometric sequence, where every term besides $a_1$ and $a_n$ cancel out and yield

$$a_1 \times \frac{1 - r^n}{1 - r}.$$

But I don't know really know how to proceed at this point, since I can't find a common ratio or difference.

$\endgroup$
5
  • 14
    $\begingroup$ Don't assume that the series must be geometric. Not all infinite series are so. $\endgroup$ Jul 24, 2015 at 3:49
  • 9
    $\begingroup$ This is a telescoping series. $\endgroup$
    – Jared
    Jul 24, 2015 at 3:55
  • 6
    $\begingroup$ "So the ratio isn't constant" - then it's not a geometric series (you proved that). "...the difference between 2 consecutive terms wasn't constant either" - so it's not an arithmetic series either (you proved that). $\endgroup$
    – Jared
    Jul 24, 2015 at 4:02
  • 2
    $\begingroup$ When in doubt, try out a few terms to see if you can find patterns. $\endgroup$
    – anon
    Jul 24, 2015 at 6:44
  • 2
    $\begingroup$ See also: Infinite Series $\sum 1/(n(n+1))$ $\endgroup$ Jul 24, 2015 at 8:58

5 Answers 5

35
$\begingroup$

$$\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right) +\cdots +\left(\frac{1}{n} - \frac{1}{n+1}\right) $$ $$ = 1-\frac12+\frac12-\frac13+\frac13-\frac14+\frac14 -\cdots +\frac{1}{n} - \frac{1}{n+1}$$ $$ = 1-\left(\frac12-\frac12\right)-\left(\frac13-\frac13\right)-\left(\frac14-\frac14\right)-\space \cdots \space - \left(\frac{1}{n}-\frac{1}{n}\right)-\frac{1}{n+1}$$

Notice how each of the terms in parentheses is zero, so we are left with: $$\boxed{\text{Sum of first n terms: }1-\frac{1}{n+1}}$$

If we want the infinite sum we must take the limit as $n \to \infty$ because $n$ is the number of terms in the sequence. So as $n$ becomes arbitrarily large, $\dfrac{1}{n}$ tends towards $0$ so we get that the sequence of finite sums approaches: $$1-0 = \boxed{1}$$


Not all infinite series need to be arithmetic or geometric! This special one is called a telescoping series.

$\endgroup$
0
6
$\begingroup$

This is not a geometric series, this is a telescoping series.

$$a_{n}=\left(\frac{1}{n}-\frac{1}{n+1}\right)$$

Note that

$$\sum_{j=1}^{n}a_{j}=1-\frac{1}{n+1}$$

You can begin to see the pattern with

$$\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)=1-\frac{1}{3}$$

Now taking the limit of the sequence of finite sums

$$\lim_{n\rightarrow\infty}\sum_{j=1}^{n}a_{j}=1$$

$\endgroup$
5
$\begingroup$

Firstly, it matters what you consider a 'term' in this series. Let's use consider that the elements in parenthesis constitute the terms.

You are given the sequence: $\{1-\tfrac 1 2, \tfrac 1 2-\tfrac 1 3, \tfrac 1 3-\tfrac 1 4, \ldots, \tfrac 1 n-\tfrac 1{n+1},\ldots\}$

This is equivalently $\{\tfrac 1 2, \tfrac 1{6}, \tfrac 1 {12}, \ldots, \tfrac 1{n(n+1)}, \ldots\}$

So the partial sum is: $S_n = \sum_{k=1}^n \tfrac 1 {k(k+1)} = \tfrac n{n+1}$

$$\{S_n\} = \{\tfrac 1 2, \tfrac 2{3}, \tfrac 3{4}, \ldots, \tfrac n{n+1}, \ldots\}$$

Can you evaluate $S_\infty = \lim_{n\to\infty} S_n$ ?


PS: Also notice that $$\require{cancel} 1 \cancel{-\tfrac 1 2+\tfrac 12} \cancel{- \tfrac 1 3 + \tfrac 1 3} \cancel{- \tfrac 1 4 + \tfrac 14} - \ldots = 1$$

$\endgroup$
0
4
$\begingroup$

Tee general term $a_k$ is $a_k=\frac{1}{k}-\frac{1}{k+1}$. So, for example, for $k=1$ the first term is $a_1=\frac{1}{1}-\frac{1}{2}$. The next term is $k=2$ is $a_2=\frac12-\frac13$. To sum the first $n$ terms, we have

$$\sum_{k=1}^n\frac{1}{k}-\frac{1}{k+1}=1-\frac{1}{n+1}$$

We can see that as $n\to \infty$, the series approaches $1$.

$\endgroup$
1
$\begingroup$

Given infinite series, $$\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots $$ The $n$th term say $T_{n}$ of the above series can be given as $$T_{n}=\left(\frac{1}{n}-\frac{1}{n+1}\right)$$ Hence, we have

  1. The sum $S_{n}$ of $n$ terms as follows $$S_{n}=\sum_{n=1}^n T_{n}$$$$=\sum_{n=1}^n\left(\frac{1}{n}-\frac{1}{n+1}\right)$$ $$=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n-1}-\frac{1}{n}\right)+\left(\frac{1}{n}-\frac{1}{n+1}\right)$$ $$=\left(1-\frac{1}{n+1}\right)=\color{blue}{\frac{n}{n+1}}$$
  2. The sum $S_{\infty}$ of infinite terms is calculated by taking the limit of $S_n$ at $n\to \infty$ as follows $$S_{\infty}=\lim_{n\to \infty}S_{n}$$ $$=\lim_{n\to \infty}\left(\frac{n}{n+1}\right)$$ $$=\lim_{n\to \infty}\left(\frac{1}{1+\frac{1}{n}}\right)$$ Now, let $\frac{1}{n}=t \implies t\to 0 \ as \ n\to \infty$ $$=\lim_{t\to 0}\left(\frac{1}{1+t}\right)$$ $$=\left(\frac{1}{1+0}\right)=\color{blue}{1}$$
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .