sum of convergent series help!

[djdavis2k]

Find the sum of the convergent series given below.

a.) sigma notation from (k=1 to infinity) of ((1/2^k)-(1/2^(k+1))

b.) sigma notation from (k=2 to infinity) of (1/((k^2)-1))

c.)sigma notation from (n=5 to infinity) of (e/pi)^(n-1)

d.) sigma notation from (i=1 to infinity) of (5^(3*i))*(7^(1-i))

for a.) u can write 2^(k+1) as (2^k)(2) so this becomes (1/2k)*(1-1/2)= 1/2^k(1/2)= i believe this approaches zero.. but i don't know how to calcuate the sum

for b.) i believe this approaches zero as well... but i'm not sure how to calculate the sum.. i believe the total sum would be equivalent to 1

for c.) i have no clue how to do this need a lot help here

for d.) i have no clue how to do this need a lot help here as well

please help me with these problems

 

[galactus]

Find the sum of the convergent series given below.

a.)\(\displaystyle \sum_{k=1}^{\infty}\left[\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right]\)

Rewrite:

\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{2^{k}}-\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{2^{k}}\)

These are geometric series with common term 1/2

\(\displaystyle \frac{\frac{1}{2}}{1-\frac{1}{2}}-\frac{1}{2}\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}}\right)=\frac{1}{2}\)

If you rewrite these, they will appear easier.

d.) \(\displaystyle \sum_{i=1}^{\infty}5^{3*i}\cdot 7^{1-i}\)

Rewrite:

\(\displaystyle 7\sum_{i=1}^{\infty}\left(\frac{125}{7}\right)^{i}\)

Since the denominator is smaller than the numerator, we can see it is infinite.

 

[djdavis2k]

Find the sum of the convergent series given below.

a.)\(\displaystyle \sum_{k=1}^{\infty}\left[\frac{1}{2^{k}}-\frac{1}{2^{k+1}}\right]\)

Rewrite:

\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{2^{k}}-\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{2^{k}}\)

These are geometric series with common term 1/2

\(\displaystyle \frac{\frac{1}{2}}{1-\frac{1}{2}}-\frac{1}{2}\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}}\right)=\frac{1}{2}\)

If you rewrite these, they will appear easier.

[quote:1ggvjel8]d.) \(\displaystyle \sum_{i=1}^{\infty}5^{3*i}\cdot 7^{1-i}\)

Rewrite:

\(\displaystyle 7\sum_{i=1}^{\infty}\left(\frac{125}{7}\right)^{i}\)

Since the denominator is smaller than the numerator, we can see it is infinite.[/quote:1ggvjel8]


Thanks for your help with a.) and d.) I understand it is a geometric series since it is in the form of a*r^n-1 and how d.) can be simplified and since the num. is larger than the denom. it goes to infinity that makes sense...

can explain to me how i would be able to do b. and c. sorry for the trouble.. thank for your help so far...

 

[galactus]

For b, note the difference of two squares.

Rewrite as \(\displaystyle \frac{1}{2(k-1)}-\frac{1}{2(k+1)}\)

Now, perhaps you can see better what it'll be.

If you list out the terms, you can see a telescoping sum.

\(\displaystyle \left(\frac{1}{2}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{6}-\frac{1}{10}\right)+\left(\frac{1}{8}-\frac{1}{12}\right)+\left(\frac{1}{10}-\frac{1}{14}\right)+.....\)

They all cancel except a few terms and that is your sum.

These things aren't that bad if you practice a little.


Try the one with e.