This example is due to Kakutani (cf. [Ric, p. 282]). Let H be a separable Hilbert space with orthonormal basis (f_m )_{m Î N}. Define a_m = e^-k for m = 2^k(2l + 1) ,  k,l = 0, 1, ¼ and also the operator T by Tf_m = a_m f_m+1 , m Î N.
Then ||T || = sup_{m Î N} | a_m | , Tⁿf_m = a_m a_m+1 ¼a_m+n-1f_m+n and so ||Tⁿ || = sup_{m Î N} (a_m a_m+1 ¼a_m+n-1).
Moreover, by the definition of the a_m, we have a₁ a₂ ¼a_2^t-1 = Õ_{j = 1}^t-1exp(-j2^t-j-1).
Therefore (a₁ a₂ ¼a_2^t-1)^{[ 1/( 2t-1)]} > (Õ_{j = 1}^t-1exp [ [(-j)/(2^j+1)]])² and if s = å_{j = 1}^¥ [j/(2^j+1)], then e^{-2 s} £ lim_n ||Tⁿ ||^{[ 1/ n]}. So T is not topologically nilpotent.
Next define the operator T_k by

Tkfm = ì í î 0 m = 2k(2l + 1), l = 0, 1, ¼ am fm+1 otherwise

Then T_k is nilpotent. But

(T - Tk)fm = ì í î e-kfm+1 m = 2k(2l + 1), l = 0, 1, ¼ 0 otherwise

Thus ||T_k - T || = e^-k, hence lim_k T_k = T in the norm topology on B(H).
Ref.
[Ric] C.E. Rickart, General theory of Banach algebras, Princeton, Van Nastrand, 1960.