On a problem book solution I was faced with the following step:
$$\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$$
I identified $\frac{1}{n+1}+\frac{1}{(n+1)^2}...$ as a geometric series so the sum would be $\frac{1}{1-r}$ so that $\frac{1}{1-\frac{1}{n+1}}=1+\frac{1}{n}$. I do not understand what I am doing worng.
Question:
What am I doing wrong?
Thanks in advance!
On
The geometric sequence does not start with $1$.
The general form is $$ a + ar + ar^2 + ... = \frac {a}{1-r}$$ for $|r|<1$
Therefor the sum
$$\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$$
is correct.
On
If you know that $a+ar+ar^2+ar^3+\cdots = \dfrac{a}{1-r}$ for $|r|\lt 1$
then here you have $a= \dfrac1{n+1}$ and $r= \dfrac1{n+1}$
so the sum is $\dfrac{\frac1{n+1}}{{1-\frac1{n+1}}}=\dfrac{1}{n+1-1}=\dfrac{1}{n}$
On
How I recommend approaching these problems:
Use the fact that if $|x| < 1$, $$1 + x + x^2 + \cdots = \dfrac{1}{1-x}\text{.} \tag{*}$$
Then, factor to rewrite the problem so that it is in terms of the equation (*).
Observe that if I factor out $\dfrac{1}{n+1}$ that
$$\dfrac{1}{n+1} + \dfrac{1}{(n+1)^2} + \cdots = \dfrac{1}{n+1}\left(1+\dfrac{1}{n+1}+\dfrac{1}{(n+1)^2}+\cdots \right)\text{.}\tag{A}$$
The sum $$1+\dfrac{1}{n+1}+\dfrac{1}{(n+1)^2} + \cdots = \dfrac{1}{1-\frac{1}{n+1}} = \dfrac{1}{(n+1-1)/(n+1)} = \dfrac{n+1}{n}\tag{B}$$ as long as $$\left|\dfrac{1}{n+1} \right| < 1 \implies |n+1| > 1 \implies n+1 > 1 \text{ and } -(n+1) < -1 \implies n > 0\text{.}$$ Thus, as long as $n > 0$, combine (A) and (B) to get $$\dfrac{1}{n+1} \cdot \dfrac{n+1}{n} = \dfrac{1}{n}$$ as desired.
Please be sure to note that this is only true when $n > 0$. I have seen mistakes in the past when people ignore assumptions made when summing infinite geometric series.
On
I identified 1n+1+1(n+1)2... as a geometric series so the sum would be 11−r so that 11−1n+1=1+1n. I do not understand what I am doing worng.
Take a look at your indexes.
A) You have $\frac 1{1+n} + (\frac 1{n+1})^2 + ..... = \sum_{k= 1}^{\infty} (\frac 1{1+n})^k$
B) So you did: Let $r = \frac 1{n+1}$ and $\sum_{k=0}^{\infty}( \frac 1{1+n})^k =\sum_{k=0}^{\infty}r^k=\frac 1{1-r} = 1 + \frac 1n$.
Take a closer look do you see the difference between $\sum_{k= 1}^{\infty} (\frac 1{1+n})^k$ in A) and $\sum_{k=0}^{\infty} (\frac 1{1+n})^k$ in B)?
....
Now $\sum_{k= 1}^{\infty} \frac 1{1+n}^k = [\sum_{k=0}^{\infty} (\frac 1{1+n})^k] - (\frac 1{n+1})^0 = ....???....$
You are using the formula $$ 1+x+x^2+x^3+\dotsb=\frac{1}{1-x}\quad (|x|<1)\quad (\star) $$ where $x=\frac{1}{n+1}$. The problem is that your desired sum omits the initial term of $1$ in $(\star)$. Hence subtract $1$ from your sum, $1+n^{-1}$, to get the right result.