how the lim (ln(2+n)/n)/(1/n) = lim (n/2+n) * (n-(2+n)/n^2 ) / (-1/n^2)....[ lim is going to + infinity]

Question

anonymous · Answer

find the value of c if $$\sum_{2}^{\infty}$$(1+c)^ -n = 2

myname · Answer

$$\sum_{2}^{\infty}1/(1+c)^n =\sum_{0}^{\infty}1/(1+c)^n - \sum_{0}^{2}1/(1+c)^n---(1)$$
$$\sum_{0}^{\infty}1/(1+c)^n = 1/1-(1/(1+c))$$
[BY USING THE FORMULA TO FIND THE SUM OF THE GEOMETRIC SERIES-->a/1-r ]

Then we simplify,
$$\sum_{0}^{\infty}1/(1+c)^n = (1+c)/c$$
Next,
$$\sum_{0}^{2}1/(1+c)^n=1+(1/1+c) = (2+c)/(1+c)$$Now we can plug in the values of these series in (1).
$$2 = ((1+c)/c) - (2+c)/(1+c)$$
Now, take the  common denominator.
$$2= ((1+c)^2-2c-c^2)/c(1+c)$$
$$2c +2c^2= 1+2c+c^2-2c-c^2$$
$$2c+2c^2-1=0$$
Here , you can use the quadratic formula to solve for c.[c is a variable like x]
By doing that you would get c =0.366 and c= -1.366
You can go head and plug the two c values in the series , and you will find which one gives you 2 and which one does not.
In this case c is 0.366