To verify if the sequence is convergent, we need to find the limit as n approaches infinity. The nth term of the sequence is given by 1/n + 1/2n + 1/3n + ... + 1/n^2. This is a harmonic series of the form 1/n, which is known to be divergent. However, in this case, each term is divided by an additional factor of n, which could potentially make the series convergent. To find the limit of the sequence as n approaches infinity, we can use the properties of limits. The limit of a sum is the sum of the limits, so we can find the limit of each term individually and then add them up. The limit as n approaches infinity of 1/n is 0. The limit as n approaches infinity of 1/2n is also 0. This pattern continues for all the terms in the sequence, so the limit of the entire sequence as n approaches infinity is 0. Therefore, the sequence is convergent, and its limit is 0.

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To verify if the sequence is convergent, we need to find the limit as n approaches infinity. The nth term of the sequence is given by 1/n + 1/2n + 1/3n + ... + 1/n^2. This is a harmonic series of the form 1/n, which is known to be divergent. However, in this case, each term is divided by an additional factor of n, which could potentially make the series convergent. To find the limit of the sequence as n approaches infinity, we can use the properties of limits. The limit of a sum is the sum of the limits, so we can find the limit of each term individually and then add them up. The limit as n approaches infinity of 1/n is 0. The limit as n approaches infinity of 1/2n is also 0. This pattern continues for all the terms in the sequence, so the limit of the entire sequence as n approaches infinity is 0. Therefore, the sequence is convergent, and its limit is 0.

If the nth term of the sequence < an > is given by1n + 12n + 13n + · · + 1n2Verify if < an > is convergent. Also find the limit of sequence.

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