Question

Write the first three therms of the following series.Please help

Answer 1

Do you have the series pattern given?

Answer 2

Yes

Answer 3

I will write it now

Answer 4

\[\sum_{1}^{\infty} \frac{ 3n - 2 }{ n ^{2} + 1 }\]

Answer 5

A series is nothing more than a definable sum of terms, and it usually follows a pattern, and it is written inside the Sigma.
\[\sum_{n=1}^{\infty} \frac{ 3n-2 }{ n^2+1 }\]
we can find any term by limiting the sum to the desired term, in this case, it's 3:
\[\sum_{n=1}^{n=3}\frac{ 3n-2 }{ n^2+1 }\]
And, this just translates to:
\[\sum_{n=1}^{n=3}\frac{ 3n-2 }{ n^2+1 }=\frac{ 3(1)-2 }{ (1)^2+1 }+\frac{ 3(2)-2 }{ (2)^2+1 } + \frac{ 3(3)-2 }{ (3)^2+1 }\]

Answer 6

srry cant u just take a limit and divide the whole thing with the biggest power which in this case is n^2

Answer 7

is it right what he jsut said?

Answer 8

i am not sure if i am right i am pretty sure he is right

Answer 9

You take the limit when you desire to know if the series is convergent or divergent.
When you want to find terms, you replace them on the actual pattern.

Answer 10

I got another question

Answer 11

Post it here or make open another topic?

Answer 12

Post it on another topic, so it doesn't get messy.

Answer 13

ok

Answer 14

oh ok srry i didnt read the question about three terms