Calculus question...first to correctly answer and give steps will get the 10pts?

Question

Here is the website.... http://www.ms.uky.edu/~qye/exam3b.pdf

and its question number 5. I know what the answer is, but I have no clue how to get it.

Thanks

Like I said, I know the answer, but I need someone to tell me how to get it.

Thanks Dr. D.

I have one question though. You stated ...
"So summation = 3 + 3/n + 4
= 7 + 7/n"

not for sure, but shouldn't it be 3 + 3/n + 4 = 7 + 3/n ? But of course since n is going to infinity the 3/n continues to get smaller and smaller. Thanks. I will make sure to give you the 10 pts.

Answer 1

7

Let's do it piece by piece, and remember k is the variable in the summation, not n.

∑ 6k/n^2 = 6/n^2 ∑ k
= 6/n^2 * (1 + 2 + 3 + ... + n)
= 6/n^2 * n ( n+1)/2
= 3*(n^2 + n) / n^2 = 3 + 3/n

∑ 4/n = 4/n + 4/n + n times
= 4

So summation = 3 + 3/n + 4
= 7 + 3/n

As n --> ∞, limit = 7

*EDIT*
Right, that was a typo and I corrected it.

Answer 2

Let's break the summation in 2 and I'll just write ∑
for ∑(k=1..n).
Let's do the second sum first:
∑ 4/n.
Since the variable of summation is k, we can pull
the 4/n outside the summation sign and we are
left with 4/n∑1 = 4n/n = 4. So the limit as n →∞ of
the second sum is 4.
Now let's go to work on the first sum:
Again, we can pull the 6/n² outside the summation sign
and get 6/n² * ∑k.
But if you remember your elementary sums, ∑k = n(n+1)/2
So we get
lim(n →∞) 6/(2n²) *n(n+1), which equals 3.
Thus the limit as n→∞ of both sums is 7.

Answer 3

easy b