limn→∞∑k=1n(2+3kn−−−−−−√4⋅3n)=
To find the value of the limit, we can rewrite the summation expression as a function of n:
S(n) = ∑(k=1 to n) (2 + 3k√(4•3n))/n
Let’s simplify the expression inside the summation first:
(2 + 3k√(4•3n))/n
= (2 + 3k√(12n))/n
= (2 + 3k√(4•3•n))/n
= (2 + 3k•2√(3n))/n
= (2 + 6k√(3n))/n
Now, substitute this simplified expression back into the original summation:
S(n) = ∑(k=1 to n) (2 + 6k√(3n))/n
Next, let’s split the summation into two parts:
S1(n) = ∑(k=1 to n) 2/n
S2(n) = ∑(k=1 to n) (6k√(3n))/n
The limit of S1(n) as n approaches infinity is straightforward:
lim(n→∞) S1(n) = lim(n→∞) ∑(k=1 to n) 2/n
Since the upper limit of the summation is ‘n’ and the value inside the summation is a constant (2/n), the sum of these terms becomes:
lim(n→∞) S1(n) = lim(n→∞) (2/n + 2/n + 2/n +
To find the value of the limit, we can rewrite the summation expression as a function of n:
S(n) = ∑(k=1 to n) (2 + 3k√(4•3n))/n
Let’s simplify the expression inside the summation first:
(2 + 3k√(4•3n))/n
= (2 + 3k√(12n))/n
= (2 + 3k√(4•3•n))/n
= (2 + 3k•2√(3n))/n
= (2 + 6k√(3n))/n
Now, substitute this simplified expression back into the original summation:
S(n) = ∑(k=1 to n) (2 + 6k√(3n))/n
Next, let’s split the summation into two parts:
S1(n) = ∑(k=1 to n) 2/n
S2(n) = ∑(k=1 to n) (6k√(3n))/n
The limit of S1(n) as n approaches infinity is straightforward:
lim(n→∞) S1(n) = lim(n→∞) ∑(k=1 to n) 2/n
Since the upper limit of the summation is ‘n’ and the value inside the summation is a constant (2/n), the sum of these terms becomes:
lim(n→∞) S1(n) = lim(n→∞) (2/n + 2/n + 2/n + … + 2/n) (n terms)
= lim(n→∞) 2 = 2
So, S1(n) approaches 2 as n approaches infinity.
Now let’s simplify S2(n):
S2(n) = ∑(k=1 to n) (6k√(3n))/n
We can move constants out of the summation:
S2(n) = (6/n) ∑(k=1 to n) (k√(3n))
= (6/n) • √(3n) ∑(k=1 to n) k
The sum of k from 1 to n can be expressed as:
∑(k=1 to n) k = n(n+1)/2
So, we can substitute this back into S2(n):
S2(n) = (6/n) • √(3n) • (n(n+1)/2)
= 3√(3) • (n+1)/√(n)
Now, let’s find the limit of S2(n) as n approaches infinity:
lim(n→∞) S2(n) = lim(n→∞) 3√(3) • (n+1)/√(n)
To apply the limit, we divide both the numerator and denominator by the highest power of n:
lim(n→∞) S2(n) = lim(n→∞) 3√(3) • (1 + 1/n)/√(1/n)
Applying the limit as n approaches infinity, 1/n approaches 0:
lim(n→∞) S2(n) = 3√(3) • (1 + 0)/√(0)
= 3√(3)/√(0)
Since the denominator approaches 0 as n approaches infinity, we need to apply L’Hôpital’s Rule to evaluate the limit of the denominator:
lim(n→∞) √(1/n) = lim(n→∞) 1/(√n)
Using L’Hôpital’s Rule:
lim(n→∞) 1/(√n) = lim(n→∞) (d/dn 1)/(d/dn √n)
= lim(n→∞) 0.5/n^2
As n approaches infinity, 0.5/n^2 approaches 0. Therefore:
lim(n→∞) √(1/n) = 0
Now, substitute this back into the previous expression:
lim(n→∞) S2(n) = 3√(3)/0
Since the denominator approaches 0, we can conclude that the limit of S2(n) does not exist.
Finally, let’s find the overall limit by adding S1(n) and S2(n):
lim(n→∞) S(n) = lim(n→∞) [S1(n) + S2(n)]
= 2 + (limit undefined)
The overall limit is undefined.
Therefore, the limit of the given expression as n approaches infinity is undefined.
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