# Math Help - nth term for series

1. ## nth term for series

Hey all!

I needed some help in how to get the nth term (Un) for some series, that are:

1/1 + 3/2 + 5/2^2 + 7/2^3 + ...

1/2 + 2/3 + 3/4 + 4/5 + ...

1/5 + 2/6 + 2^2/7 + 2^3/8 + 2^4/9 + ...

I know the nth term (Un) for each but I wanted to know how you get them.

Thank you

Hey all!

I needed some help in how to get the nth term (Un) for some series, that are:

1/1 + 3/2 + 5/2^2 + 7/2^3 + ...

1/2 + 2/3 + 3/4 + 4/5 + ...

1/5 + 2/6 + 2^2/7 + 2^3/8 + 2^4/9 + ...

I know the nth term (Un) for each but I wanted to know how you get them.

Thank you
Basicaly by inspection, and making some assumptions.

Look at the first:

1/1 + 3/2 + 5/2^2 + 7/2^3 + ...

The numerators are consecutive odd numbers, which we may write as 2k-1
for k=1, 2, .. The denominators are consecutive powers of 2 which we may
write 2^(k-1) k=1, 2, .. (the -1 in the exponent is required here as I am
taking the first term to correspond to k=1 for both the numerator and
denominator), so the k-th term is (2k-1)/2^{k-1}.

Now look at the second:

1/2 + 2/3 + 3/4 + 4/5 + ...

The numerators are consecutive integers which we may write k, k=1, 2, ..
The denominators are also consecutive integers, but starting from 2, so
may be written (k+1), k=1, 2, .. Hence the general term is k/(k+1).

RonL

These have pretty obvious patterns . . .

1/1 + 3/2 + 5/2² + 7/2³ + ...

The numerators are: 1, 3, 5, 7, ...
. . The general numerator is: 2n - 1

The denominators are: 2^
0, 2^1, 2^2, 2^3, ...
. . The general denominator is: 2^
{n-1}

The general term is: .a
n .= .(2n - 1) / 2^{n-1}

1/2 + 2/3 + 3/4 + 4/5 + ...

The numerators are: 1, 2, 3, 4, ...
. . The general numerator is: n

The denominators are: 2, 3, 4, 5, ...
. . The general denominator is: n + 1

The general term is: .a
n .= .n / (n + 1)

1/5 + 2/6 + 2²/7 + 2³/8 + ...

The numerators are: 2^
0, 2^1, 2^2, 2^3, ...
. . The general numerator is: 2^{n-1}

The denominators are 5, 6, 7, 8, ...
. . The general denominator is: n + 4

The general term is: .a
n .= .2^{n-1} / (n + 4)

4. ## thanks

Thank you!

Yeah does seem quite straight forward now. I was getting confused on the 2^(n-1) and was forgetting the 2^0, 2^1 terms.

Thanks again

Kind regards,