Evaluating the Limit of a Series with Radical Expressions as n Approaches Infinity

limn→∞∑k=1n(2+3kn−−−−−−√4⋅3n)=

To evaluate the limit of the given series as n approaches infinity, let’s break it down step by step

To evaluate the limit of the given series as n approaches infinity, let’s break it down step by step.

We have the series ∑(k=1 to n) (2 + 3k√(4⋅3n)).

First, let’s simplify the expression inside the square root: 4⋅3n = 12n.

Now, the expression becomes ∑(k=1 to n) (2 + 3k√(12n)).

Next, let’s focus on the term inside the square root and simplify it further. We can rewrite √(12n) as √(4⋅3⋅n^2) = 2n√3.

The expression now becomes ∑(k=1 to n) (2 + 3k(2n√3)).

Expanding it further: ∑(k=1 to n) (2 + 6kn√3).

Now, we can factor out 2: ∑(k=1 to n) 2(1 + 3kn√3).

Finally, we can split the series into two separate parts: ∑(k=1 to n) 2 + ∑(k=1 to n) 6kn√3.

The first part, ∑(k=1 to n) 2, is simply 2 added n times, which results in 2n.

The second part, ∑(k=1 to n) 6kn√3, can be rewritten as 6√3n ∑(k=1 to n) k.

The series ∑(k=1 to n) k is a sum of consecutive positive integers and is given by the formula n(n + 1) / 2.

Substituting that into the expression, we get 6√3n * (n(n + 1) / 2).

Simplifying further: 3√3n^2 * (n + 1).

Expanding that expression: 3√3n^3 + 3√3n^2.

Now, combining the two parts: 2n + 3√3n^3 + 3√3n^2.

As n approaches infinity, the dominant term that contributes to the limit is n^3.

Therefore, the limit of the given series as n approaches infinity is ∞.

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