Calculating the Limit of a Sum with a Simplified Expression inside: An Analysis of Dominant Terms and the Limit as n Approaches Infinity

limn→∞∑k=1n(2+3kn−−−−−−√4⋅3n)=

To calculate the limit of the given sum, we can rewrite the expression inside the sum as follows:

2 + 3k√(4⋅3n)

Since we are taking the limit as n approaches infinity, we need to simplify the expression inside the square root

To calculate the limit of the given sum, we can rewrite the expression inside the sum as follows:

2 + 3k√(4⋅3n)

Since we are taking the limit as n approaches infinity, we need to simplify the expression inside the square root. We can rewrite √(4⋅3n) as √4 ⋅ √(3n), which gives us:

2 + 3k√(4⋅3n) = 2 + 3k√4 ⋅ √(3n) = 2 + 3k⋅2⋅√(3n)

Now we can substitute this simplified expression back into the original sum:

∑(k=1 to n)[2 + 3k√(4⋅3n)] = ∑(k=1 to n)[2 + 3k⋅2⋅√(3n)]

We can split this sum into two separate terms:

∑(k=1 to n)2 + ∑(k=1 to n)[3k⋅2⋅√(3n)]

The first term, ∑(k=1 to n)2, is simply a sum of the number 2 repeated n times, so it becomes:

∑(k=1 to n)2 = 2 + 2 + 2 + … (n times) = 2n

The second term, ∑(k=1 to n)[3k⋅2⋅√(3n)], can be simplified further. We can factor out the constants 3 and 2 from the sum, giving us:

3⋅2⋅∑(k=1 to n)[k⋅√(3n)]

Now, let’s focus on the remaining sum ∑(k=1 to n)[k⋅√(3n)]. We can expand it as follows:

k⋅√(3n) = √(k^2)⋅√(3n) = √(3n)⋅√(k^2) = √(3n)⋅k

So, the sum becomes:

∑(k=1 to n)[k⋅√(3n)] = ∑(k=1 to n)[√(3n)⋅k] = √(3n)⋅∑(k=1 to n)k

The sum ∑(k=1 to n)k is a well-known arithmetic series and can be found using the formula:

∑(k=1 to n)k = n(n+1)/2

Substituting this back into our expression:

∑(k=1 to n)[k⋅√(3n)] = √(3n)⋅[n(n+1)/2] = (n(n+1)/2)⋅√(3n)

Now, let’s revisit the original expression:

∑(k=1 to n)[2 + 3k√(4⋅3n)] = 2n + (n(n+1)/2)⋅√(3n)

As n approaches infinity, we can see that the dominant term in this expression is (n(n+1)/2)⋅√(3n). The other term, 2n, is negligible compared to this dominant term.

Therefore, the limit as n approaches infinity is:

lim(n→∞) [∑(k=1 to n)[2 + 3k√(4⋅3n)]] = lim(n→∞) [(n(n+1)/2)⋅√(3n)]

By using the limit properties and simplifying further, we can find the final answer.

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