# Calculate lmit from this:a(n)= f(1)+f(2)+.....+f(n) - ln [n(n+1)/2].And function is defined from - infinity to -2, open interval, f(x)= ln [1+(2/x)].

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You need to evaluate the sum `f(1)+f(2)+...+f(n)` such that:

Notice that the function is `f(x) = ln[1+(2/x)], ` hence, bringing the terms to a common denominator yields:

`f(x) = ln((x+2)/x) = ln(x+2) - ln x`

You need to substitute 1 for x such that:

`f(1) = ln(1+2) - ln 1`

`f(1) = ln 3 - ln 1`

You need to substitute 2 for x such that:

`f(2) = ln4 - ln 2`

You need to substitute 3 for x such that:

`f(3) = ln 5 - ln 3`

You need to substitute 4 for x such that:

`f(4) = ln6 - ln 4`

You need to substitute 5 for x such that:

`f(5) = ln7 - ln 5`

You need to add `f(1)+f(2)+...+f(n)` such that:

`ln 3 - ln 1 + ln 4 - ln 2 + ln 5 - ln 3 + ln 6 - ln 4 + ln 7 - ln 5 + ... + ln(n) - ln(n-2) + ln(n+1) - ln(n-1) + ln(n+2) - ln(n)`

Reducing like terms yields:

`ln(n+2) - ln(2)`

Hence, substituting`ln(n+2) - ln(2) for f(1)+f(2)+...+f(n` ) yields:

`a(n) = ln(n+2) + ln(n+1)- ln(2) - ln(n(n+1)/2)`

`a(n) = ln (((n+1)(n+2))/2)/(n(n+1)/2)`

`a(n) = ln ((n+2)/n)`

You need to evaluate the limit of a(n) such that:

`lim_(n-gtoo) a(n) = lim_(n-gtoo)ln ((n+2)/n)`

`ln lim_(n-gtoo) (n(1+2/n))/n`

Reducing by n yields:

`ln lim_(n-gtoo) (1+2/n)/1 = ln 1 `

`lim_(n-gtoo) a(n) = 0`

**Hence, evaluating the limit of a(n), under given conditions, yields `lim_(n-gtoo) a(n) = 0.` **

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