Dear friends, Please read our latest blog post for an important announcement about the website. ❤, The Socratic Team

#lim_(n->oo)(sum_(k=1)^n(sqrt(2n^2+k)/(2n^2+k)))#?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

5

This answer has been featured!

Featured answers represent the very best answers the Socratic community can create.

Learn more about featured answers

Hammer Share
Apr 1, 2018

Answer:

#lim_(n->oo) sum_(k=1)^(n) sqrt(2n^2+k)/(2n^2+k) = 1/sqrt2#

Explanation:

We have

#lim_(n->oo) sum_(k=1)^(n) sqrt(2n^2+k)/(2n^2+k) = lim_(n->oo)sum_(k=1)^(n) 1/sqrt(2n^2+k) =#

#= lim_(n->oo) sum_(k=1)^n 1/n * 1/sqrt(2+k/n^2)#

Now, let's try to remember the definition of an integral. This might seem strange, but it will come of great use.

The definite integral #int_a^b f(x)dx# represents the #color(red)("net area")# defined by the graph of the function #f(x)# and the x-axis. One way to find the value of an integral is by aproximating it with infinitely many rectangles ("boxes") of equal or not width.

This concept is called a #color(red)("Riemann Sum")#.

There are infinitely many rectangles, therefore, if there are #n# rectangles, we have to take the limit as #n->oo#.

Now, let's take the case where the width is constant between all boxes. If we define #x_1,x_2,...,x_n# to be some "marks" on the x-axis such that the width of the #i#-th rectangle is #x_(i+1) - x_i = Delta x_i#, then its lenght is #f(x_i)#. Since the width is equal, #Deltax_i# is the same for all #i#'s. We can more easily call it #color(red)(Delta#.

The Riemann Sum aproximates an integral by being the sum of all the rectangles. In order to simplify this further, let's take the particular case where #color(red)(a = 0# and #color(red)(b=1#. We have:

#int_0^1 f(x)dx = lim_(n->oo) sum_(i=1)^(n) Delta*f(x_i)#

This is only true if #x_n# is equal to #1#.

One way to make #Delta# constant is to define #x_k = k/n#, for some #k#, #1<=k<=n#. This way, #x_n = 1# and

#color(red)(Delta) = x_(i+1)-x_i = (i+1)/n-i/n = color(red)(1/n)#.

#int_0^1 f(x)dx = lim_(n->oo) sum_(i=1)^n 1/n * f(i/n)#.

This looks familiar to our original sum, #lim_(n->oo) sum_(k=1)^n 1/n * 1/sqrt(2+k/n^2)#.

All we are left to do is to assume they're equal and then solve the integral.

#lim_(n->oo) sum_(k=1)^n 1/n * 1/sqrt(2+k/n^2) = lim_(n->oo) sum_(i=1)^n 1/n * f(i/n)#

#i# and #k# are just the indexes, the name of which doesn't matter. Both sums could be counting #i# or #k#, it's irrelevant.

#lim_(n->oo) sum_(k=1)^n 1/n * 1/sqrt(2+k/n^2) = lim_(n->oo) sum_(k=1)^n 1/n * f(k/n)#

#=> f(k/n) = 1/sqrt(2+k/n^2) => f(x) = 1/sqrt(2+x/n)#.

We have to find the integral of this function:

#lim_(n->oo) int_0^1 dx/sqrt(2+x/n)#

Since #n->oo#, this means that #x/n = 0#, as #x# is finite.

#lim_(n->oo) int_0^1 dx/sqrt2 = 1/sqrt2#.

This is because it forms a rectangle with the x-axis, the width and lenght of which is #1# and #1/sqrt2#, respectively.

As a conclusion,

#color(blue)(lim_(n->oo) sum_(k=1)^(n) sqrt(2n^2+k)/(2n^2+k) = 1/sqrt2#.

Was this helpful? Let the contributor know!
1500